1 Introduction

Since the discovery of a scalar particle with 125GeV at the LHC [1, 2], there has been a strong theoretical effort and experimental search for extra scalar particles. A particularly interesting framework is provided by N Higgs doublet models (NHDM). For example, models with three Higgs doublets (3HDM) supplemented by the \(A_4\) discrete symmetry have long been explored in connection with flavour physics; see, for example, [3,4,5,6,7,8,9,10].

But, it was only in 2012 that Degee, Ivanov and Keus carefully addressed the identification of viable candidates for minima in the \(A_4\) 3HDM [11].Footnote 1 They provide conditions for only two of the solutions of the stationarity equations to be minima. Ref. [13] revisited this issue, proposing conditions for two other solutions to be minima. We have found that some conditions are incorrect, while most only guarantee that the minimum is local, not global. And, in some cases, such local minima occur for potentials which are in fact unbounded from below. Indeed, a different but related issue concerns the conditions for the potential to be bounded from below (BFB). This turns out to have an interesting impact in our analysis.

As pointed out in Ref. [14], one must be careful to distinguish conditions for the potential to be bounded from below along the directions of neutral vacua (BFB-n), from the conditions that guarantee that the potential is bounded from below along the directions of charge breaking vacua (BFB-c). It turns out that the necessary and sufficient BFB-n conditions for the \(A_4\) have been conjectured in [15] and proven in [16]. These hinged on the fact that Ref. [11] fully mapped the orbit space of the neutral directions of the \(A_4\) 3HDM. At the moment, there is no such analysis for the charge breaking (CB) directions.

Here, we address all these issues. After summarizing the known results, we point out in Sect. 2.3 that, contrary to previous assumptions, one must look at five, not four, candidate minima. In Sect. 3, we re-express the BFB-n conditions in a way that connects directly with the existence of local minima. Section 4 has the first main results. We write the correct conditions for all five points (candidates for global minima of the \(A_4\) 3HDM) to be local minima. We then show that, in general, this does not guarantee that such points are global minima, providing both the relevant conditions and explicit numerical counter-examples. We end with a summary of the necessary and sufficient conditions for a point to be the global minimum consistent with BFB-n. This assumes both that the potential will be BFB-c and that there is no lower-lying CB global minimum.

However, one must still address the issue of the charge breaking directions: Does the potential satisfy BFB-c conditions?; Are there lower-lying CB global minima? This is what we turn to in Sect. 5. We start by developing a safe sufficient condition for the potential to be BFB-c. We show that, without this condition, two dangerous situations may occur. There may be some points that pass BFB-n and are global minima of neutral directions, but which, nevertheless, are not the global minima of the potential, since there are lower-lying CB solutions. There may even be points that pass the conditions just mentioned but correspond to potentials unbounded from below along charged directions. Using extensive numerical simulations, we found that, after the conditions for BFB-n and of global minima are imposed, there is only one type of point (identified as point D below) where we found lower-lying (that is, true) CB global minima or potentials unbounded from below. We conjecture that this is a general feature. We present our conclusions in Sect. 6.

2 The \(A_4\)-symmetric 3HDM potential

2.1 Potential and candidates for global minimum

The potential for the \(A_4\) symmetric three Higgs doublet model (3HDM) is, in the notation of [11],

$$\begin{aligned} V_{A_4} =&- \frac{M_0}{\sqrt{3}} \left( \phi _1 ^ \dagger \phi _1 + \phi _2 ^ \dagger \phi _2 + \phi _3 ^ \dagger \phi _3\right) + \frac{\Lambda _0}{3} \left( \phi _1 ^ \dagger \phi _1 + \phi _2 ^ \dagger \phi _2 + \phi _3 ^ \dagger \phi _3\right) ^ 2 \nonumber \\&+ \Lambda _1 \left[ \left( \text {Re}\left\{ \phi _1^\dagger \phi _2\right\} \right) ^2 + \left( \text {Re}\left\{ \phi _2^\dagger \phi _3\right\} \right) ^2 + \left( \text {Re}\left\{ \phi _3^\dagger \phi _1\right\} \right) ^2 \right] \nonumber \\&+ \Lambda _2 \left[ \left( \text {Im}\left\{ \phi _1^\dagger \phi _2\right\} \right) ^2 + \left( \text {Im}\left\{ \phi _2^\dagger \phi _3\right\} \right) ^2 + \left( \text {Im}\left\{ \phi _3^\dagger \phi _1\right\} \right) ^2 \right] \nonumber \\&+ \frac{\Lambda _3}{3} \left[ (\phi _1 ^ \dagger \phi _1) ^ 2 + (\phi _2 ^ \dagger \phi _2) ^ 2+ (\phi _3 ^ \dagger \phi _3) ^ 2 - (\phi _1^\dagger \phi _1) (\phi _2^\dagger \phi _2) \right. \nonumber \\&\left. - (\phi _2^\dagger \phi _2) (\phi _3^\dagger \phi _3) - (\phi _3^\dagger \phi _3) (\phi _1^\dagger \phi _1)\right] \nonumber \\&+ \Lambda _4 \left[ \text {Re}\left\{ \phi _1^\dagger \phi _2\right\} \, \text {Im}\left\{ \phi _1^\dagger \phi _2\right\} + \text {Re}\left\{ \phi _2^\dagger \phi _3\right\} \, \text {Im}\left\{ \phi _2^\dagger \phi _3\right\} \right. \nonumber \\&\left. + \text {Re}\left\{ \phi _3^\dagger \phi _1\right\} \, \text {Im}\left\{ \phi _3^\dagger \phi _1\right\} \right] \,. \end{aligned}$$
(1)

Setting \(\Lambda _4 = 0\), it reduces to the \(S_4\) invariant 3HDM potential. Therefore, the \(A_4\) and \(S_4\) models have 6 and 5 free parameters, respectively.

Contrary to what seemed implicit in [11], and became explicit in subsequent papers using their result – including, for example, [13, 17] – there are 5 (and not 4) different vev directions that can lead to a global minimum. They are:

$$\begin{aligned}&A: \tfrac{v}{\sqrt{2}}(1,0,0)\, , \, B: \tfrac{v}{\sqrt{6}}(1,1,1)\, , \, D: \tfrac{v}{2}(1,e^{i\alpha },0)\, , \nonumber \\*[2mm]&C: \tfrac{v}{\sqrt{6}}( \pm 1,\eta ,\eta ^{-1})\, , \quad C': \tfrac{v}{\sqrt{6}}(\pm 1,\omega ,\omega ^2)\, \end{aligned}$$
(2)

where \(\eta = e^{i \pi /3}\), \(\omega = \eta ^2 = e^{2i\pi /3}\), and \(\alpha \) is fixed by the relations

$$\begin{aligned}&\sin 2\alpha = - \frac{\Lambda _4}{\sqrt{\left( \Lambda _1 - \Lambda _2 \right) ^2 + \Lambda _4^2}}\, , \nonumber \\*[2mm]&\cos 2\alpha = - \frac{\Lambda _1 - \Lambda _2}{\sqrt{\left( \Lambda _1 - \Lambda _2 \right) ^2 + \Lambda _4^2}} \, . \end{aligned}$$
(3)

The minima have been normalized such that \(v_1^2 + v_2^2 + v_3^2 =v^2/2\). Notice that \(A_4\) contains \({\mathbb {Z}}_2 \times {\mathbb {Z}}_2\), and that one has invariance under global hypercharge. As a result, the potential is invariant under independent sign changes of \(\phi _1\), \(\phi _2\), and \(\phi _3\), as can be checked from direct inspection of (1). Thus, the ± in C and \(C'\) is immaterial.

Part of the confusion in the literature as to the number of minima appeared because \(C'\) is not defined in [11], although it is clear from their figure 3 that there are two distinct minima; C and \(C'\). Moreover, [17] renames \(C'\) as C and does not mention a second possibility.

Why is this relevant? It is relevant because, for a given set of \(\Lambda \)’s, C and \(C'\) do not yield the same value for the potential at that minimum. That is \(V(C) \ne V(C')\); one is the true global minimum and the other is not. Adding to the confusion, starting from C:

$$\begin{aligned} (-1, \eta , \eta ^{-1}) = \eta ^{-1} (\omega ^2,\omega ,1)\, , \end{aligned}$$
(4)

Since an overall phase is irrelevant, it would thus seem that one could get to \(C'\) simply by transposing \(\phi _1 \leftrightarrow \phi _3\). However, transpositions such as this are not in \(A_4\). Indeed, performing \(\phi _1 \leftrightarrow \phi _3\) in Eq. (1), the potential does not remain invariant. Fortunately, the transformed potential is obtained from the initial one by simply changing \(\Lambda _4 \rightarrow -\Lambda _4\). Thus, the formulae obtained in [11] for C can be adapted to \(C'\) by simply changing the sign of \(\Lambda _4\). Still, we emphasize that, for chosen \(\Lambda \)’s, \(V(C) \ne V(C')\).

Part of the confusion may be due to the fact that transpositions such as \(\phi _1 \leftrightarrow \phi _3\) are a part of \(S_4\). In that case, the potential has \(\Lambda _4=0\) and the distinction between C and \(C'\) disappears.

2.2 Vevs, mass spectra, and values of the potential

The spectra for \(A_4\) has been studied in [11], and here we summarize those results. We denote the charged scalar pairs by \(H_{i}^\pm \) (\(i=1,2\)), and the new neutral scalars (besides the lowest lying SM-like Higgs) by \(H_k\) (\(k=1,2,3,4\)). In cases where a mass ordering of the latter is clear, we denote by \(h_i\) (\(H_i\)) the masses of the lightest (heaviest) neutral scalars.

The SM-like Higgs boson, \(H_{SM}\), always has a mass given by \(m_{H_{SM}}^2 = \frac{2}{\sqrt{3}}M_0\). Thus, \(M_0>0\) for all cases. This is related to the following fact. One can use a unitary \(3 \times 3\) transformation in the space of doublets (the so-called Higgs-family space), to go into some other basis. A particularly useful basisFootnote 2 is the Higgs basis [18], where only the first doublet carries a vev [19,20,21]. In this basis, the imaginary part of the neutral component of the first doublet corresponds always to the neutral would-be Goldstone boson \(G^0\). The corresponding real part of the neutral component of the first doublet corresponds, in general, to a neutral Higgs field that will mix with other four neutral components into the physical (mass eigenstate) neutral scalars. However, in the particular case of the 3HDM with an exact \(A_4\) symmetry, for each of the five points of the \(A_4\) 3HDM parameter space identified in [11] as candidates for global minima, there is always a transformation into a Higgs basis where that neutral scalar decouples completely from the remaining four fields. Of course, this would not be true for minima of a general 3HDM (in which case, even in the Higgs basis, the five neutral fields other than \(G^0\) mix through a \(5 \times 5\) mass matrix); but it is true for the case discussed in [11].

3 \(\mathbf {A:\ (1, 0, 0)}\)

For the (1, 0, 0) solution of the stationarity condition, the magnitude of the vev is given by

$$\begin{aligned} v^2 = \frac{\sqrt{3}M_0}{\Lambda _0 + \Lambda _3}. \end{aligned}$$
(5)

We will sometimes use \(\Lambda _A=\Lambda _3\). The spectra is then expressed as

$$\begin{aligned} m_{H^\pm _i}^2&= - \frac{v^2}{2} \Lambda _3 \qquad \text {(doubly degenerate)}\, , \end{aligned}$$
(6)
$$\begin{aligned} m_{h_i}^2&= \frac{v^2}{4} \left( \Lambda _1 + \Lambda _2 - 2 \Lambda _3 - \sqrt{(\Lambda _1 - \Lambda _2)^2 + \Lambda _4^2} \right) \nonumber \\&\text {(doubly degenerate)}\, , \end{aligned}$$
(7)
$$\begin{aligned} m_{H_i}^2&= \frac{v^2}{4} \left( \Lambda _1 + \Lambda _2 - 2 \Lambda _3 + \sqrt{(\Lambda _1 - \Lambda _2)^2 + \Lambda _4^2} \right) \qquad \nonumber \\&\text {(doubly degenerate)}\, . \end{aligned}$$
(8)

The value of the potential at the minimum is

$$\begin{aligned} V(A) = -\frac{M_0^2}{4(\Lambda _0 + \Lambda _3)}\, . \end{aligned}$$
(9)

4 \(\mathbf {B:\ (1, 1, 1)}\)

For the (1, 1, 1) solution of the stationarity condition, the magnitude of the vev is given by

$$\begin{aligned} v^2 = \frac{\sqrt{3}M_0}{\Lambda _0 + \Lambda _1}. \end{aligned}$$
(10)

We will sometimes use \(\Lambda _B = \Lambda _1\). The charged Higgs spectra has less degeneracies than at point A, and we find

$$\begin{aligned} m_{H^\pm _i}^2&= \frac{v^2}{2} \left( - \Lambda _1 \pm \frac{\Lambda _4}{2 \sqrt{3}}\right) \, , \end{aligned}$$
(11)
$$\begin{aligned} m_{h_i}^2&= \frac{v^2}{12} \bigg ( - 5 \Lambda _1 + 3 \Lambda _2 + 2 \Lambda _3 - \nonumber \\&\quad - \sqrt{(- \Lambda _1 + 3 \Lambda _2 - 2 \Lambda _3)^2 + 12 \Lambda _4^2} \bigg ) \nonumber \\&\text {(doubly degenerate)}\, , \end{aligned}$$
(12)
$$\begin{aligned} m_{H_i}^2&= \frac{v^2}{12} \bigg ( - 5 \Lambda _1 + 3 \Lambda _2 + 2 \Lambda _3 + \nonumber \end{aligned}$$
(13)
$$\begin{aligned}&\quad + \sqrt{(- \Lambda _1 + 3 \Lambda _2 - 2 \Lambda _3)^2 + 12 \Lambda _4^2} \bigg ) \nonumber \\&\text {(doubly degenerate)}\, . \end{aligned}$$
(14)

The value of the potential at the minimum is

$$\begin{aligned} V(B) = -\frac{M_0^2}{4(\Lambda _0 + \Lambda _1)}\, . \end{aligned}$$
(15)

5 \(\mathbf {C:\ (1, \eta , \eta ^{-1})}\)

For the \((1, \eta , \eta ^{-1})\) solution of the stationarity condition, the magnitude of the vev is given by

$$\begin{aligned} v^2 = \frac{\sqrt{3} M_0}{\Lambda _0 + \Lambda _C}, \end{aligned}$$
(16)

where

$$\begin{aligned} \Lambda _C= \frac{1}{4} \left( \Lambda _1 + 3 \Lambda _2 + \sqrt{3} \Lambda _4 \right) \, . \end{aligned}$$
(17)

The masses are given by

$$\begin{aligned} m_{H_i^\pm } =&-\frac{v^2}{2} \left( \Lambda _2 +\frac{1}{2\sqrt{3}} \Lambda _4\right) \, , \nonumber \,\\&-\frac{v^2}{4} \left( \Lambda _1 + \Lambda _2 +\frac{2}{\sqrt{3}} \Lambda _4\right) \, . \end{aligned}$$
(18)
$$\begin{aligned} m_{h_i}^2&= \frac{v^2}{12}(A - B)\, , \quad m_{H_i}^2 = \frac{v^2}{12}(A + B) \nonumber \\&\qquad \qquad \text {(doubly degenerate)}\, , \end{aligned}$$
(19)

where

$$\begin{aligned} A= & {} -8\Lambda _C + 3(\Lambda _1 + \Lambda _2) + 2 \Lambda _3 \nonumber \\= & {} \Lambda _1 - 3 \Lambda _2 + 2 \Lambda _3 - 2 \sqrt{3} \Lambda _4 \nonumber \\= & {} (a+c)+(b+c)\, , \end{aligned}$$
(20)
$$\begin{aligned} B= & {} \bigg [(4\Lambda _C - 3(\Lambda _1 + \Lambda _2) + 2 \Lambda _3)^2 \nonumber \\&+ 4 (2\Lambda _C + \Lambda _1 -3 \Lambda _2)^2\bigg ]^{1/2} \nonumber \\*[1mm]= & {} \bigg [(-2 \Lambda _1 + 2 \Lambda _3 + \sqrt{3} \Lambda _4)^2 \nonumber \\&+ (3 \Lambda _1 - 3 \Lambda _2 + \sqrt{3} \Lambda _4)^2\bigg ]^{1/2} \nonumber \\= & {} \sqrt{(a-c)^2+(b-c)^2}\, , \end{aligned}$$
(21)

andFootnote 3

$$\begin{aligned} a=3(\Lambda _1 - \Lambda _2)\, , \ \ \ b= 2(\Lambda _3 - \Lambda _1)\, , \ \ \ c=- \sqrt{3} \Lambda _4\, . \end{aligned}$$
(22)

The value of the potential at the minimum is

$$\begin{aligned} V(C) = -\frac{M_0^2}{4(\Lambda _0 + \Lambda _C)}\, . \end{aligned}$$
(23)

6 \(\mathbf {C':\ (1, \omega , \omega ^2)}\)

This case can be obtained from the previous one through \(\Lambda _4 \rightarrow - \Lambda _4\). The vev becomes,

$$\begin{aligned} v^2 = \frac{\sqrt{3} M_0}{\Lambda _0 + \Lambda _{C'}}, \end{aligned}$$
(24)

where

$$\begin{aligned} \Lambda _{C'}= \frac{1}{4} \left( \Lambda _1 + 3 \Lambda _2 - \sqrt{3} \Lambda _4 \right) \, . \end{aligned}$$
(25)

The charged Higgs masses are given by

$$\begin{aligned} m_{H_i^\pm }&= -\frac{v^2}{2} \left( \Lambda _2 - \frac{1}{2\sqrt{3}} \Lambda _4\right) \, , \nonumber \\&\quad -\frac{v^2}{4} \left( \Lambda _1 + \Lambda _2 - \frac{2}{\sqrt{3}} \Lambda _4\right) \, . \end{aligned}$$
(26)
$$\begin{aligned} m_{h_i}^2&= \frac{v^2}{12}(A' - B')\, , \ \ m_{H_i}^2 = \frac{v^2}{12}(A' + B') \nonumber \\&\qquad \quad \text {(doubly degenerate)}\, , \end{aligned}$$
(27)

where

$$\begin{aligned} A'= & {} -8 \Lambda _{C'} + 3(\Lambda _1 + \Lambda _2) + 2 \Lambda _3\, , \end{aligned}$$
(28)
$$\begin{aligned} B'= & {} \bigg [(4 \Lambda _{C'} - 3(\Lambda _1 + \Lambda _2) + 2 \Lambda _3)^2 \nonumber \\&+ 4 (2 \Lambda _{C'} + \Lambda _1 -3 \Lambda _2)^2\bigg ]^{1/2}\, . \end{aligned}$$
(29)

The value of the potential at the minimum is

$$\begin{aligned} V(C') = -\frac{M_0^2}{4(\Lambda _0 + \Lambda _{C'})}\, . \end{aligned}$$
(30)

7 \(\mathbf {D:\ (1, e^{i\alpha }, 0)}\)

For the \((1, e^{i\alpha }, 0)\) solution of the stationarity condition, the magnitude of the vev is given by

$$\begin{aligned} v^2 = \frac{4 \sqrt{3} M_0}{4 \Lambda _0 + \Lambda _3 - 3 \tilde{\Lambda }} = \frac{\sqrt{3} M_0}{\Lambda _0 + \Lambda _D}, \end{aligned}$$
(31)

where we have defined

$$\begin{aligned} \tilde{\Lambda } = \frac{1}{2} \left( \sqrt{(\Lambda _1 - \Lambda _2)^2 + \Lambda _4^2} - (\Lambda _1 + \Lambda _2)\right) \, , \end{aligned}$$
(32)

and

$$\begin{aligned} \Lambda _D= & {} \frac{1}{4}(\Lambda _3 - 3 \tilde{\Lambda }) \qquad \ \ \ \ \ \ \ \nonumber \\= & {} \frac{3}{8} \big [ \Lambda _1 + \Lambda _2 + \frac{2}{3} \Lambda _3 \nonumber \\&-\sqrt{(\Lambda _1 - \Lambda _2)^2 + \Lambda _4^2} \big ]. \end{aligned}$$
(33)

The corresponding spectra is

$$\begin{aligned} m_{H^\pm _i}^2&= \frac{v^2 }{2} \tilde{\Lambda }, \quad \frac{v^2}{4} (\tilde{\Lambda } - \Lambda _3)\, , \end{aligned}$$
(34)
$$\begin{aligned} m_{H_i}^2 =&\frac{v^2}{2} ( \tilde{\Lambda } + \Lambda _3)\, , \quad \frac{v^2}{2}(2 \tilde{\Lambda } + \Lambda _1 + \Lambda _2)\, , \end{aligned}$$
(35)
$$\begin{aligned}&\frac{v^2}{4} \bigg [ (\tilde{\Lambda } + \Lambda _1 + \Lambda _2 - \Lambda _3) \nonumber \\&\pm (2 \tilde{\Lambda } + 3 \Lambda _1 - \Lambda _2)\, \sqrt{\frac{\tilde{\Lambda } + \Lambda _2}{ 2 \tilde{\Lambda } + \Lambda _1 + \Lambda _2}} \bigg ]\, . \end{aligned}$$
(36)

Notice that the definition of \(\tilde{\Lambda }\) in Eq. (31) guarantees that \(\tilde{\Lambda } + \Lambda _1 >0\), \(\tilde{\Lambda } + \Lambda _2 >0\), and that \(2 \tilde{\Lambda } + \Lambda _1 + \Lambda _2 \ge \mid \Lambda _1-\Lambda _2 \mid > 0 \) . This implies that the second Eq. (34) is always positive. In this case there are no degeneracies in the expressions for the masses of the neutral scalars, and we have separated by commas the four \(m_{H_i}^2\). The value of the potential at the minimum is

$$\begin{aligned} V(D) = -\frac{M_0^2}{4\Lambda _0 + \Lambda _3 - 3 \tilde{\Lambda }}\, . \end{aligned}$$
(37)

Alternatively, we may write

$$\begin{aligned} V(D) = -\frac{M_0^2}{4(\Lambda _0 + \Lambda _D)}\, . \end{aligned}$$
(38)

7.1 The necessary distinction between C and \(C'\)

As mentioned in the introduction, the five minima A, B, D, C, and \(C'\) appear in figure 3 of [11]. But the corresponding \(C'\) masses and conditions are not presented and the fact that the two minima are distinct is absent from [17] , and subsequent literature, including [13]. Imagine that one is looking at a particular point in parameter space, corresponding to the values

$$\begin{aligned} M_0= & {} 13.887500 v^2, \quad \Lambda _0 = 2.928220\, , \nonumber \\ \Lambda _1= & {} -0.305014, \quad \Lambda _2 = -1.018350\, , \nonumber \\ \Lambda _3= & {} 1.365140, \quad \Lambda _4 = -1.053270\, . \end{aligned}$$
(39)

After some preliminary numerical calculations, one might suspect that \(C': (1, \omega , \omega ^2)\) (unfortunately renamed C in [13, 17]) is the lowest minimum, with the value of the potential at the minimum given by Eq. (29) as

$$\begin{aligned} V(C') = -0.189506 \times 10^2 v^4\, . \end{aligned}$$
(40)

But, it is not. The correct minimum is \(C: (1,\eta , \eta ^{-1})\) and the value of the potential at the minimum is not \(V(C')\) but, rather, given by Eq. (22) as

$$\begin{aligned} V(C) = -0.295417 \times 10^2 v^4\, . \end{aligned}$$
(41)

It is true, that given the \(\Lambda _4 \rightarrow - \Lambda _4\) interchange between C and \(C'\), the roles of C and \(C'\) will be reversed in the case

$$\begin{aligned} M_0= & {} 13.887500 v^2\, , \quad \Lambda _0 = 2.928220\, , \nonumber \\ \Lambda _1= & {} -0.305014\, , \quad \Lambda _2 = -1.018350\, , \nonumber \\ \Lambda _3= & {} 1.365140\, , \quad \Lambda _4 = +1.053270\, . \end{aligned}$$
(42)

But that is not the issue. The problem with ignoring C (or \(C'\), as the case may be) is that, for a given set of parameters of the potential, the two candidate vacua give different values for the potential; one of the vacua is the global minimum; the other is not. This issue is even more troublesome if one is looking for the minima in a completely softly-broken \(A_4\) potential, where such a parameter symmetry is absent, and one is trying to use the results of the exact \(A_4\)-symmetric potential as a guide.

7.2 Comparison with the literature

The expressions for the vevs and masses in cases A, B, D, and C are contained in Ref. [11]. We agree with their results, except that the \(\Lambda _4\) term in their Eq. (B.10) – their Eq. (43) on the arxiv version – for the charged Higgs masses in case C must be changed from \(\sqrt{3} \Lambda _4/2\) into \(\Lambda _4/(2 \sqrt{3})\) in order to agree with our Eq. (17).

The expressions for the vevs and masses in cases A, B, D, and \(C'\) are repeated in Ref. [13], but with the notational change that \(C'\) is renamed C and the appropriate change is made in the sign in front of \(\Lambda _4\). We have found a number of misprints in their Eqs. (12), (23), and (24).

The conditions that guarantee that a given vacuum is indeed the absolute minimum are discussed in the next section. It turns out that the conditions identified in [11, 13] to guarantee that B is the absolute minimum are necessary but not sufficient to that effect. In addition, equations to guarantee that \(C'\) and D are the global minimum are not presented in [11]. They are presented in [13], where they are wrong. This is what we turn to next.

8 Bounded from below conditions (neutral directions)

The full conditions that guarantee that the exact \(A_4\) symmetric 3HDM potential is bounded from below (BFB) are not known. The problem is that the form of the orbit space for the charged solutions (solutions of the vev that break electric charge) is not known. And, as illustrated in [14, 15], guaranteeing BFB only along the neutral directions does not guarantee that the potential is indeed BFB.

The conditions for BFB of the \(A_4\) 3HDM along the neutral directions (BFB-n) have been conjectured in [15], and proved to hold in [16]. They are:

$$\begin{aligned}&\Lambda _0 + \Lambda _3 \ge 0\, , \end{aligned}$$
(43)
$$\begin{aligned}&\frac{4}{3}(\Lambda _0 + \Lambda _3) + \frac{1}{2} (\Lambda _1 + \Lambda _2) - \Lambda _3 \nonumber \\&\quad - \frac{1}{2} \sqrt{(\Lambda _1 - \Lambda _2)^2+\Lambda _4^2} \ge 0\, , \end{aligned}$$
(44)
$$\begin{aligned}&\Lambda _0 + \frac{1}{2} (\Lambda _1 + \Lambda _2) + \frac{1}{2} (\Lambda _1 - \Lambda _2) \cos {(2 k \pi /3 )}\nonumber \\&\quad + \frac{1}{2} \Lambda _4 \sin {(2 k \pi /3 )} \ge 0\, \ \ (k=1,2,3)\, . \end{aligned}$$
(45)

But, the left hand side of Eq. (43) is equal to \(4(\Lambda _0+\Lambda _D)/3\). Thus, the second BFB condition may be traded for

$$\begin{aligned} \Lambda _0+\Lambda _D \ge 0\, , \end{aligned}$$
(46)

which holds if and only if Eq. (30) representing \(v^2\) at point D is positive. Similarly, Eq. (44) with \(k=3\) may be written as

$$\begin{aligned} \Lambda _0+\Lambda _1 \ge 0\, , \end{aligned}$$
(47)

which according to Eq. (10) holds if and only if the expression for \(v^2\) at point B is positive. Finally, Eq. (44) with \(k=1\) and \(k=2\) may be written, respectively, as

$$\begin{aligned}&\Lambda _0+ \Lambda _C \ge 0\, , \nonumber \\&\Lambda _0 + \Lambda _{C'} \ge 0\, , \end{aligned}$$
(48)

which according to Eqs. (15) and (23) hold if and only if the expressions for \(v^2\) at points C and \(C'\) are positive, respectively. Said otherwise, the BFB-n conditions of Ref. [16] may be written compactly as

$$\begin{aligned} \Lambda _0 + \Lambda _X \ge 0 \ \ \ \ (\text {for all }\, X=A,B,C,C',D)\, . \end{aligned}$$
(49)

These equations guarantee that the right hand side of equations (5), (10), (15), (23), (30) representing \(v^2\) at points X are all positive.

To be specific, when solving the stationarity equations, one finds the trivial (0, 0, 0) solution. One also finds possible solutions at points \(P=A,B,C,C',D\), whenever the right hand side of

$$\begin{aligned} v^2 = \frac{\sqrt{3}M_0}{\Lambda _0 + \Lambda _P} \end{aligned}$$
(50)

is positive. Of course, the \(\Lambda \) parameters might be such that \(\Lambda _0 + \Lambda _C >0\), while \(\Lambda _0 + \Lambda _{C'}<0\). In that case, C is a critical point but \(C'\) is not. When we mention in the text that we require \(v^2 >0\) at a given point P, what we mean is that we require \(\Lambda _0 + \Lambda _P >0\). Equation (48) can be taken to mean that the potential is BFB-n if and only if all points \(P=A,B,C,C',D\) are critical points simultaneously. The physical explanation is discussed in detail in [15]; it hinges on the fact that, for the exact \(A_4\) potential, there is a connection between the BFB conditions and the value of the quartic part of the potential at the candidates for global minima.

9 Conditions for global minimum

Recall that requiring that the 125 GeV scalar has a positive mass squared implies that \(M_0>0\) for all cases. Let us assume that we found that a given point P is a local minimum. Then, in a notation that should be obvious from Eqs. (9), (14), (36), (22), and (29), we conclude that, at that point P one must have \(\Lambda _0 + \Lambda _P > 0\). Let us continue to imagine that P is a local minimum. We wish to know whether V(X) could be lower at some other point \(X \ne P\). But

$$\begin{aligned} V(X) - V(P) = \frac{M_0^2}{4(\Lambda _0+\Lambda _P)(\Lambda _0 + \Lambda _X)} \left[ \Lambda _X - \Lambda _P \right] \, . \nonumber \\ \end{aligned}$$
(51)

The prefactor on the right hand side of this equation is guaranteed to be positive by the BFB conditions in Eq. (48). Thus, \(V(X) < V(P)\) can only be attained if it is possible to have

$$\begin{aligned} \Lambda _X - \Lambda _P < 0\, , \end{aligned}$$
(52)

together with \(\Lambda _0 + \Lambda _P > 0\), the other conditions needed for P to be a local minimum, and the conditions for BFB-n potential.Footnote 4

10 \(\mathbf {A:\ (1, 0, 0)}\)

Using the geometrical minimization technique, Ref. [11] identifies the conditions for (1, 0, 0) to be “stable” asFootnote 5

$$\begin{aligned} \Lambda _3&< 0\, , \qquad \Lambda _0> \mid \Lambda _3\mid > -\Lambda _2, -\Lambda _1\, , \nonumber \\ \Lambda _4^2&< 4(\Lambda _1 + \mid \Lambda _3\mid )(\Lambda _2 + \mid \Lambda _3\mid )\, . \end{aligned}$$
(53)

It is interesting that (52) can be re-obtained in the following way. Requiring that \(m_{H_i^\pm }^2\) in (6) is positive, we obtain \(\Lambda _3=-\mid \Lambda _3\mid \) is negative. Thus, from (5), \(\Lambda _0 > \mid \Lambda _3\mid \). Adding (7) and (8), we conclude that \(\mid \Lambda _3\mid >-(\Lambda _1 + \Lambda _2)/2\). Requiring that \(m_{h_i}^2\) in (7) is positive implies that \((\Lambda _1 + \Lambda _2 + 2 \mid \Lambda _3\mid )^2 > (\Lambda _1 - \Lambda _2)^2+\Lambda _4^2\), from which \(\Lambda _4^2 < 4(\Lambda _1 + \mid \Lambda _3\mid )(\Lambda _2 + \mid \Lambda _3\mid )\). We wish for \(\Lambda ^2_4 \ne 0\); \(\Lambda _4=0\) would lead to the \(S_4\) symmetric 3HDM. This coincides almost exactly with (52). Defining

$$\begin{aligned} y_1 = \Lambda _1 + \mid \Lambda _3\mid \, , \ \ \ y_2 = \Lambda _2 + \mid \Lambda _3\mid \, , \end{aligned}$$
(54)

the previous conditions require \(y_1 + y_2 \equiv \Sigma > 0\) and \(y_1 y_2 > 0\). But this forces both \(y_1\) and \(y_2\) be larger than 0 (and, although irrelevant to our argument, smaller than \(\Sigma \), related through (7) and (8) with \(m_{h_i}^2 + m_{H_i}^2\)). Thus, \(\mid \Lambda _3\mid > -\Lambda _2, -\Lambda _1\), and we have recovered all the conditions in (52). A curious fact is that in this argument we did not have to require that V(A) lie lower than the values of the potential at the other solutions of the stationarity equation; V(B), V(D), V(C), \(V(C')\). Indeed, one can show that requiring that, at the vacuum A, \(v^2>0\) and that all squared masses are positiveFootnote 6guarantees that the potential is BFB-n and that \(V(A) < V(\text {other})\). Said otherwise, the conditions (52), which guarantee that A is a local minimum, also guarantee that it is the global minimum.

We start by proving that (52) imply that the potential is BFB-n. To this end, we define

$$\begin{aligned} x_k = \Lambda _0 + \Lambda _k\ \ \ (k=1,2,3)\, . \end{aligned}$$
(55)

Equation (52) may be re-expressed as

$$\begin{aligned}&\Lambda _3< 0, x_{1,2,3}>0, x_3< x_1,\, x_2\, , \nonumber \\&\Lambda _4^2 < 4(x_1 - x_3)(x_2 - x_3)\, . \end{aligned}$$
(56)

The second equation (55) implies that (42) and (46) hold. We now turn to Eq. (47). They are satisfied if

$$\begin{aligned} 4(\Lambda _0 + \Lambda _C) = x_1 + 3 x_2 \pm \sqrt{3} \Lambda _4 > 0 \end{aligned}$$
(57)

is positive. But, using the last Eq. (55),

$$\begin{aligned}&x_1 + 3 x_2 - \sqrt{3} \mid \Lambda _4\mid \nonumber \\&\quad> x_1 + 3 x_2 - 2 \sqrt{3} \sqrt{(x_1-x_3)(x_2-x_3)} \nonumber \\&\quad = (x_1-x_3)+ 3(x_1-x_3) - 2 \sqrt{3(x_1-x_3)(x_2-x_3)} + 4 x_3 \nonumber \\&\quad = \left( \sqrt{x_1-x_3} - \sqrt{3(x_2-x_3)} \right) ^2 + 4 x_3 > 0\, , \end{aligned}$$
(58)

implying that Eq. (47) holds. We now turn to Eq. (45). We use the last Eq. (55) to show that

$$\begin{aligned} (x_1-x_2)^2 + \Lambda _4^2= & {} \left[ (x_1-x_3)-(x_2-x_3)\right] ^2 + \Lambda _4^2 \nonumber \\< & {} \left[ (x_1-x_3)-(x_2-x_3)\right] ^2 + 4(x_1-x_2)(x_1-x_3) \nonumber \\= & {} \left[ (x_1-x_3)+(x_2-x_3)\right] ^2\, . \end{aligned}$$
(59)

As a result

$$\begin{aligned}&\left( x_1 + x_2 + \frac{2}{3} x_3\right) ^2 - \left[ (x_1-x_2)^2 + \Lambda _4^2 \right] \nonumber \\&\quad>\left( x_1 + x_2 + \frac{2}{3} x_3\right) ^2 - \left( x_1+x_2-2x_3\right) ^2 >0\, , \end{aligned}$$
(60)

leading to

$$\begin{aligned} \frac{8}{3}(\Lambda _0 + \Lambda _D) = x_1 + x_2 + \frac{2}{3} x_3 - \sqrt{(x_1-x_2)^2 + \Lambda _4^2} >0,\nonumber \\ \end{aligned}$$
(61)

and Eq. (45) is verified. We have thus proved that the conditions (52) imply automatically that the potential is BFB-n.

We now wish to prove that conditions (52) imply that \(\Lambda _{B,C,C',D} > \Lambda _A\), which, according to (50) means that A is the global minimum. In fact \(\Lambda _B -\Lambda _A = \Lambda _1 + \mid \Lambda _3\mid > 0\) is precisely one of Eq. (52). Using Eqs. (16), (24), and (53), we find

$$\begin{aligned} 4(\Lambda _{C,C'} - \Lambda _A)= & {} \Lambda _1 + 3 \Lambda _2 -4\Lambda _3 \pm \sqrt{3}\Lambda _4 \nonumber \\= & {} y_1 + 3 y_2 \pm \sqrt{3} \Lambda _4\, . \end{aligned}$$
(62)

We now use the last Eq. (52) to find

$$\begin{aligned} (y_1 + 3 y_2)^2 - 3 \mid \Lambda _4\mid ^2&> (y_1 + 3 y_2)^2 - 12 y_1 y_2 \nonumber \\&= (y_1 - 3 y_2)^2 >0 \, . \end{aligned}$$
(63)

Thus, conditions (52) imply \(\Lambda _{C,C'} > \Lambda _A\). Similarly

$$\begin{aligned} \frac{8}{3}(\Lambda _D - \Lambda _A) = y_1 + y_2 - \sqrt{(y_1-y_2)^2 + \Lambda _4^2}\, , \end{aligned}$$
(64)

and

$$\begin{aligned} (y_1+y_2)^2 - (y_1-y_2)^2 -\Lambda _4^2 = 4y_1 y_2 - \Lambda _4^2 >0\, , \end{aligned}$$
(65)

where the last inequality follows from the last Eq. (52). This completes the proof that asking that A be a local minimum, as in Eq. (52), automatically guarantees that the potential is BFB-n and that A is also the global minimum.

We will prove next that this very simple and favorable situation does not occur for the remaining points of interest.

11 \(\mathbf {B:\ (1, 1, 1)}\)

Using the geometrical minimization technique, Ref. [11] identifies the conditions for (1, 1, 1) to be “stable” asFootnote 7

$$\begin{aligned}&\Lambda _1< 0, \Lambda _0> \mid \Lambda _1\mid > -\Lambda _2, -\Lambda _3, \nonumber \\&\Lambda _4^2< 12\Lambda _1^2, \Lambda _4^2 < 2(\Lambda _3 + \mid \Lambda _1\mid )(\Lambda _2 + \mid \Lambda _1\mid )\, . \end{aligned}$$
(66)

As in case A, we can recover these conditions by requiring that, at point B, \(v^2 >0\) and that all expressions for the squared masses are positive. Thus, these conditions ensure that B is a local minimum. These conditions are also enough to prove that \(V(B)< V(A)\).

However, contrary to what we saw in case A, here the conditions (65) are not enough to guarantee that \(V(B) < V(C)\). Indeed, we have shown that if a point satisfies (65) but, in addition, satisfies also

$$\begin{aligned}&\Lambda _0 + \Lambda _C> 0\, , \ \ \ \text {and}\ \ \ \Lambda _4 < 0\, , \ \ \ \nonumber \\&\text {and} \ \ \mid \Lambda _4\mid > \sqrt{3}(\Lambda _2 + \mid \Lambda _1\mid )\, , \end{aligned}$$
(67)

then it has \(V(C) < V(B)\). A numeric example is

$$\begin{aligned}&\Lambda _0 = 8.8427, \Lambda _1 = -7.36842, \Lambda _2 = -3.62124,\nonumber \\&\Lambda _3 = 2.62323, \Lambda _4 = -8.03341 < 0, \end{aligned}$$
(68)

which leads to \(\Lambda _C= -8.0366\) (and, thus, \(\Lambda _0 + \Lambda _C >0\)), and \(\mid \Lambda _4\mid -\sqrt{3}(\Lambda _2+\mid \Lambda _1\mid ) = 1.5431 > 0\). As a result, in this case,

$$\begin{aligned}&\frac{4}{M_0^2} V(B)= -0.678297\, , \ \ \ \nonumber \\&\frac{4}{M_0^2} V(C) = -1.24054 < V(B)\, , \end{aligned}$$
(69)

even tough Eq. (65) are satisfied.

Similarly, points where Eq. (65) are satisfied but also satisfy

$$\begin{aligned}&\Lambda _0 + \Lambda _{C'}> 0\, , \ \ \ \text {and}\ \ \ \Lambda _4> 0\, , \ \ \ \nonumber \\&\text {and}\ \ \ \mid \Lambda _4\mid > \sqrt{3}(\Lambda _2 + \mid \Lambda _1\mid )\, , \end{aligned}$$
(70)

lead to \(V(C') < V(B)\).

Finally, we can find points for which Eq. (65) are satisfied but still lead to \(V(B) > V(D)\). Those are points for which, in addition to (65),

$$\begin{aligned} 4 \Lambda _0> 3 \tilde{\Lambda } - \Lambda _3 > 4 \mid \Lambda _1\mid \, . \end{aligned}$$
(71)

12 \(\mathbf {C:\ (1, \eta , \eta ^{-1})}\)

Ref. [13] claims that, in the region of parameter space where

$$\begin{aligned}&\Lambda _2 < 0\, , \ \mid \Lambda _2\mid> \mid \Lambda _1\mid \, , \ \ \Lambda _3> \Lambda _1 \, , \nonumber \\&4\Lambda _0 + 3 \Lambda _1 > 3\mid \Lambda _2\mid , \ \ \text {(not correct)} \end{aligned}$$
(72)

\((1, \eta , \eta ^{-1})\) is the global minimum.Footnote 8 This is incorrect. We find that these conditions are also not the correct necessary and sufficient conditions for C to be a local minimum.

To find the correct equations for C to be a local minimum, we proceed as follows. The first Eq. (17) implies that

$$\begin{aligned} - \sqrt{3} \Lambda _4 = c> 6 \Lambda _2\, . \end{aligned}$$
(73)

The second Eq. (17) implies that

$$\begin{aligned} - \sqrt{3} \Lambda _4 = c> \frac{3}{2} (\Lambda _1 + \Lambda _2)\, . \end{aligned}$$
(74)

The last two equations imply that

$$\begin{aligned} \Lambda _C < 0\, , \end{aligned}$$
(75)

a result which can also be obtained by adding the first and second equations (17). For Eq. (15) to yield a positive vev squared, we must have

$$\begin{aligned} \Lambda _0 > - \Lambda _C = \mid \Lambda _C\mid . \end{aligned}$$
(76)

Adding the two expressions in Eq. (18), we find

$$\begin{aligned} 2c > - (a+b) = - (\Lambda _1 - 3 \Lambda _2 + 2 \Lambda _3) \equiv 2 c_{ab}\, . \end{aligned}$$
(77)

For the first equation (18) to yield a positive mass squared, we also need that \(A^2 > B^2\). After some calculations, this requires that

$$\begin{aligned} c < c_- \, \ \ \ \text {or}\ \ \ c > c_+ \, , \end{aligned}$$
(78)

where

$$\begin{aligned} 2 c_\pm = -3(a+b) \pm \sqrt{9(a+b)^2-4ab}\, . \end{aligned}$$
(79)

However, one can show that

$$\begin{aligned} c_-< c_{ab} < c_+\, . \end{aligned}$$
(80)

Thus, the solution \(c<c_-\) is impossible, and, moreover, the simultaneous requirement that \(c>c_{ab}\) and \(c>c_+\) reduces to

$$\begin{aligned} c > c_+\ \ \ \text {always}\, . \end{aligned}$$
(81)

The conditions for C to be a local minimum are, thus,

$$\begin{aligned}&\Lambda _0> \mid \Lambda _C\mid \, , \ \ \ - \sqrt{3} \Lambda _4 = c> 6 \Lambda _2\, , \nonumber \\&c> \frac{3}{2} (\Lambda _1 + \Lambda _2)\, , \ \ \ c > c_+\, . \end{aligned}$$
(82)

We point out that, even after imposing (81), \(\Lambda _1\), \(\Lambda _2\), and \(\Lambda _4\) can appear with either sign.

It is true that, under the conditions that A be the global minimum in (52), it is guaranteed that \(V(A) < V(C)\). So, if C is to be the global minimum, one or more of the conditions in (52) must be violated. We are now interested in what happens to \(V(A)-V(C)\) under the conditions (81), that, as we have just seen, guarantee that V(C) is a local minimum. Requiring \(V(A) < V(C)\) is only possible if \(\Lambda _0+\Lambda _3>0\), \(\Lambda _3 <0\), and \(\Lambda _C - \Lambda _3 >0\). But the latter condition forces \(c < -(a+2b)\), which is impossible given that \(c>c_+\). We conclude that, under conditions (81), \(V(C) < V(A)\) always.

In contrast, requiring \(V(B) < V(C)\) is possible if \(\Lambda _0+\Lambda _1>0\), \(\Lambda _1 <0\), and \(\Lambda _C - \Lambda _1 >0\). The latter condition forces \(c < -a\), which is only compatible with \(c>c_+\) when \(a<0\) and \(a+b >0\). Such points are difficult to generate in a blind scan. One point obeying (81) but having \(V(B) < V(C)\) is

$$\begin{aligned}&\Lambda _0 = 2.302450\, , \ \ \ \Lambda _1 = -1.937900\, , \ \ \ \nonumber \\&\qquad \qquad \Lambda _2 = -0.877449\, , \nonumber \\&\Lambda _3 = 1.487450\, , \ \ \ \Lambda _4 = -1.424600\, , \end{aligned}$$
(83)

which leads to \(V(B)-V(C)= -0.901551 \times (M_0^2/4)\).

We now compare V(C) with \(V(C')\). Given that \(\Lambda _{C'} = \Lambda _C + c/2\),

$$\begin{aligned} V(C')-V(C) = \frac{M_0^2}{8} \frac{1}{\Lambda _0 + \Lambda _C} \left[ \frac{c}{\Lambda _0 + \Lambda _{C'}} \right] \, . \end{aligned}$$
(84)

But, using the BFB conditions in Eq. (48), we know that \(\Lambda _0 + \Lambda _{C}>0\) and \(\Lambda _0 + \Lambda _{C'}>0\). We conclude that, under conditions (81),

$$\begin{aligned} c> 0\Rightarrow & {} V(C') > V(C)\, , \\ c< 0\Rightarrow & {} V(C') < V(C)\, . \end{aligned}$$
(85)

We now turn to V(C) versus V(D). Following (51), for \(V(D) < V(C)\) we would need \(\Lambda _0 + \Lambda _D > 0 \) and \(\Lambda _D - \Lambda _C < 0\), which, since \(\Lambda _C < 0\), requires \(\Lambda _D < 0\) and \(\Lambda _0> \mid \Lambda _D\mid > \mid \Lambda _C\mid \). In addition, one can show that

$$\begin{aligned} 8(\Lambda _D - \Lambda _C) = A - \sqrt{a^2 + 3 c^2}\, , \end{aligned}$$
(86)

which would be negative for \(V(D) < V(C)\), despite \( A = a + b + 2c\) defined in (19) being positive. We find numerically that these conditions are possible. An example is,

$$\begin{aligned}&\Lambda _0 = 2.695860\, , \ \ \ \Lambda _1 = 2.176150\, , \ \ \ \nonumber \\&\qquad \qquad \Lambda _2 = -1.581800\, , \nonumber \\&\Lambda _3 = 0.585923\, , \ \ \ \Lambda _4 = -0.856958\, , \end{aligned}$$
(87)

which leads to \(V(D)-V(C)= -0.0229917 \times (M_0^2/4)\).

13 \(\mathbf {C':\ (1, \omega , \omega ^2)}\)

The discussion here parallels that on C. One may merely define \(c'=+\sqrt{3} \Lambda _4\), performing the substitution \(c \rightarrow c'\) in the formulae of case C. Recall, however, that, given a numerical point, C and \(C'\) are not “equivalent”. Indeed, for a given point with \(\Lambda _4 \ne 0\), V(C) is always different from \(V(C')\).

Point \(C'\) here is what is named point C in Ref. [13], where it is claimed that the conditions in (71) are necessary and sufficient for \(C'\) to be a global minimum. We have shown that, for example, the point

$$\begin{aligned}&\Lambda _0 = 1.079130\, , \ \ \ \Lambda _1 = 2.436530\, , \ \ \ \nonumber \\&\qquad \qquad \Lambda _2 = 0.626603\, , \nonumber \\&\Lambda _3 = 0.823911\, , \ \ \ \Lambda _4 = 2.721530\, , \end{aligned}$$
(88)

corresponds to a potential which is bounded from below, where \(C'\) is both the local and global minimum and which (in particular because it has \(\Lambda _2>0\)) violates (71).

14 \(\mathbf {D:\ (1, e^{i\alpha }, 0)}\)

Ref. [13] claims that, in the region of parameter space where

$$\begin{aligned}&\Lambda _2 < 0\, , \ \ \mid \Lambda _2\mid> \mid \Lambda _3\mid \, \ \ \Lambda _1> \Lambda _3 \nonumber \\&4\Lambda _0 + 3 \Lambda _3 > 3\mid \Lambda _2\mid , \ \ \text {(not correct)} \end{aligned}$$
(89)

\((1, e^{i\alpha }, 0)\) is the global minimum. This is incorrect. We find that these conditions are also not the correct necessary and sufficient conditions for D to be a local minimum. Indeed, for example, the point

$$\begin{aligned}&\Lambda _0 = 1.178820\, , \ \ \ \Lambda _1 = -0.137928\, , \ \ \ \nonumber \\&\qquad \qquad \Lambda _2 = 1.458710\, , \nonumber \\&\Lambda _3 = -0.607344\, , \ \ \ \Lambda _4 = -2.660920\, , \end{aligned}$$
(90)

corresponds to a potential which is bounded from below, where D is both the local and global minimum and which (in particular because it has \(\Lambda _2>0\)) violates the conditions (89).

Conversely, the point

$$\begin{aligned}&\Lambda _0 = 1.676240\, , \ \ \ \Lambda _1 = 0.145391\, , \ \ \ \nonumber \\&\qquad \qquad \Lambda _2 = -1.943450\, , \nonumber \\&\Lambda _3 = -0.044656\, , \ \ \ \Lambda _4 = -0.358779\, , \end{aligned}$$
(91)

passes all conditions in (89), but it is the point C (not the point D) which is the global minimum of the potential.

The correct necessary and sufficient conditions for the point D to be a local minimum can be cumbersome to writeFootnote 9. To simplify their appearance, we introduce the notation

$$\begin{aligned} \ell _k = \tilde{\Lambda } + \Lambda _k\ \ \ (k=1,2,3)\, . \end{aligned}$$
(92)

Recall that from the definition of \(\tilde{\Lambda }\) in Eq. (31), \(\ell _1 > 0\), \(\ell _2 > 0\) and, thus, the second Eq. (34) is automatically satisfied. The first Eq. (33) requires \(\tilde{\Lambda }>0\), the second Eq. (33) requires \(\ell _3 < 2 \tilde{\Lambda }\), and the first Eq. (34) requires \(\ell _3>0\). Eq. (35) may be rewritten as

$$\begin{aligned} \frac{4}{v^2} m_{H_i}^2 = (\ell _1 + \ell _2 - \ell _3) \pm (3 \ell _1 - \ell _2) \sqrt{\frac{\ell _2}{\ell _1 + \ell _2}}\, . \end{aligned}$$
(93)

Requiring that these masses squared be positive imposes \(\ell _3<L\), where

$$\begin{aligned} L = (\ell _1 + \ell _2) - \mid 3 \ell _1 - \ell _2\mid \sqrt{\frac{\ell _2}{\ell _1 + \ell _2}}\, , \end{aligned}$$
(94)

and the right hand side is always positive, as one can easily check. Finally, one must ensure that \(\Lambda _0 + \Lambda _D >0\). As a result, The conditions for D to be a local minimum are, thus,

$$\begin{aligned}&\tilde{\Lambda }>0\, , \ \ \ 0< \ell _3< 2 \tilde{\Lambda }\, , \ \ \ \nonumber \\&\ell _3 < L\, \ \ \ \Lambda _0 > \tilde{\Lambda } - \frac{\ell _3}{4}\, . \end{aligned}$$
(95)

One practical form to simulate points in this approach is to choose \(\Lambda _1\), \(\Lambda _2\), and \(\Lambda _4\) such that the first equation holds. This determines \(\ell _1\) and \(\ell _2\), to substitute in Eq. (94). Then, choose some \(\ell _3\) (which determines \(\Lambda _3\)) satisfying the next two equations, and finally taking \(\Lambda _0\) to obey the last equation.

14.1 Neutral global minima: summary

We are now ready to generate points guaranteed to be the local minimum in \(A_4\) symmetric potentials, which are BFB (at least along the neutral directions). The Procedure is as follows:

  1. 1.

    Chose some point P from the list of candidates provided by [11]: \(\{A,B,C,C',D\}\). For a given point, use the respective conditions for local minimum. These are presented in Eq. (52) for point A, in Eq. (65) for point B, in Eq. (81) for point C, in Eq. (81) with \(\Lambda _4 \rightarrow - \Lambda _4\) for point \(C'\), and in Eq. (95) for point D.

  2. 2.

    Impose the BFB-n conditions as in Eq. (48).

  3. 3.

    Guarantee that the chosen point is indeed the global minimum by imposing the conditions \(\Lambda _X -\Lambda _P > 0\), for all \(X \ne P\).

This procedure guarantees that all points correspond to potentials that are BFB-n. But they do not guarantee that the potential is BFB along the charge breaking directions (BFB-c). This is what we discuss in the next section.

15 Protecting against global minima along charged directions

15.1 A lower bound for BFB-c

Using the gauge freedom, one can write the most general vacua of any 3HDM as [14]

$$\begin{aligned}&\langle \phi _1 \rangle = \sqrt{r_1}\, \begin{pmatrix} 0\\ *[1mm] 1 \end{pmatrix}\, , \quad \langle \phi _2 \rangle =\sqrt{r_2}\, \begin{pmatrix} \sin (\alpha _2)\\ *[1mm] \cos (\alpha _2) e^{i \beta _2} \end{pmatrix},\nonumber \\&\langle \phi _3 \rangle =\sqrt{r_3} e^{i \gamma }\, \begin{pmatrix} \sin (\alpha _3)\\ *[1mm] \cos (\alpha _3) e^{i \beta _3} \end{pmatrix}. \end{aligned}$$
(96)

So far, we have discussed what happens to the potential (evaluated at the vacuumFootnote 10) along the neutral directions \(\alpha _2=\alpha _3=0\).

But, as shown in [14, 15], a potential can be bounded from below along the neutral directions (BFB-n), but not along the charge breaking directions (BFB-c). Unfortunately, the BFB-c conditions are not yet known for the \(A_4\) 3HDM. So, in this section we develop sufficient conditions for BFB-c, following the technique developed in [22].

It is convenient to define [14]

$$\begin{aligned} z_{ij}= (\phi _i^\dagger \phi _i) (\phi _j^\dagger \phi _j) - (\phi _i^\dagger \phi _j) (\phi _j^\dagger \phi _i) \quad \text {(no sum)}\, . \end{aligned}$$
(97)

When evaluated at the vevs, it is always true that [14]

$$\begin{aligned} 0\le z_{ij} \le r_i r_j . \end{aligned}$$
(98)

It is also useful to follow the notation of Ivanov and Faro [14], adapted to the \(A_4\) 3HDM, and write the quartic potential as

$$\begin{aligned} V_4= V_N + V_{CB} + V_{A_4}\, , \end{aligned}$$
(99)

where

$$\begin{aligned} V_N&=\frac{a}{2}\left[ (\phi _1^\dagger \phi _1)^2 + (\phi _2^\dagger \phi _2)^2+ (\phi _3^\dagger \phi _3)^2\right] \nonumber \\&\quad + b \left[ (\phi _1^\dagger \phi _1) (\phi _2^\dagger \phi _2) + (\phi _2^\dagger \phi _2) (\phi _3^\dagger \phi _3) + (\phi _3^\dagger \phi _3) (\phi _1^\dagger \phi _1) \right] \, , \end{aligned}$$
(100)
$$\begin{aligned} V_{CB}=c \left( z_{12}+ z_{13}+ z_{23}\right) \, , \end{aligned}$$
(101)

and

$$\begin{aligned} V_{A_4}= \frac{d}{2}\bigg \{ e^{- 2 i \alpha } \left[ \left( \phi _1^\dagger \phi _2\right) ^2 +\left( \phi _2^\dagger \phi _3\right) ^2 + \left( \phi _3^\dagger \phi _1\right) ^2 \right] + \text {h.c.} \bigg \}\, , \nonumber \\ \end{aligned}$$
(102)

where \(\alpha \) is defined by Eq. (3). The relation with (1) is

$$\begin{aligned}&a= \frac{2}{3}(\Lambda _0 + \Lambda _3),= \frac{1}{3}(2\Lambda _0 - \Lambda _3) + \frac{1}{2} (\Lambda _1 + \Lambda _2)\, , \nonumber \\&c= - \frac{1}{2} (\Lambda _1 + \Lambda _2), d= \frac{1}{2} \sqrt{(\Lambda _1 - \Lambda _2)^2 + \Lambda _4^2}\, . \end{aligned}$$
(103)

Since the problem of finding necessary and sufficient conditions for the \(A_4\) potential is not yet solved for the charged directions, we follow the technique in [22] and seek a lower potential to bound the original potential. Using the parameterization in Eq. (96), we get

$$\begin{aligned} V_{A_4}^{Lower}= -d \left( r_1 r_2 + r_1 r_3 + r_2 r_3\right) \, . \end{aligned}$$
(104)

Now we go back to \(V_{CB}\). Because of Eq. (98) we always have

$$\begin{aligned} V_{CB} \ge V_{CB}^{\mathrm{Lower}}= \min (0,c)\left( r_1 r_2 + r_1 r_3 + r_2 r_3 \right) \, . \end{aligned}$$
(105)

Now, using Eq. (105), we can finally write

$$\begin{aligned} V_4 > V_4^{\mathrm{Lower}} = V_N + V_{CB}^{\mathrm{Lower}} + V_{A_4}^{\mathrm{Lower}}\, . \end{aligned}$$
(106)

Therefore this potential bounds the original potential; if it is BFB, so it is the original one. We start by defining

$$\begin{aligned} A = \begin{pmatrix} A_d &{} A_o &{} A_o\\ A_o &{} A_d &{} A_o\\ A_o &{} A_o &{} A_d \end{pmatrix}\, , \end{aligned}$$
(107)

where the diagonal (\(A_d\)) and off-diagonal (\(A_0\)) matrix elements are given, respectively by

$$\begin{aligned} A_d= & {} a\ =\ \frac{2}{3}(\Lambda _0+\Lambda _3)\, , \nonumber \\ A_o= & {} b + \text {min}(0,c)-d \nonumber \\= & {} \frac{1}{3}(2\Lambda _0 - \Lambda _3) +\frac{1}{2}(\Lambda _1+\Lambda _2) + \nonumber \\&+ \min (0,-\frac{1}{2}(\Lambda _1+\Lambda _2)) \nonumber \\&- \frac{1}{2} {\sqrt{(\Lambda _1-\Lambda _2)^2+\Lambda _4^2}}\, . \end{aligned}$$
(108)

To check the BFB for \(V_4^{\mathrm{Lower}}\), we just have to check the copositivity of the matrix A [23, 24]. This leads to

$$\begin{aligned}&A_d \ge 0\, , \ \ \ \ A_d + A_o \ge 0\, , \ \ \ \ \nonumber \\&\sqrt{A_d} (A_d + 3 A_0) + \sqrt{2(A_d + A_o)^3} \ge 0\, . \end{aligned}$$
(109)

The addition of the last equation implies simply that \(A_o \ge - A_d/2\). Thus, the copositivity conditions are

$$\begin{aligned} A_d \ge 0\, , \ \ \ \ A_o \ge -A_d/2\, . \end{aligned}$$
(110)

Such conditions will ensure sufficient conditions for the potential to be BFB, but they are not necessary. There will be good points in parameter space that are discarded by this procedure. We will come to this issue below when we compare the respective sets of points.

15.2 Numerical simulations and conjectures

We have made a very large number of simulations of parameters obeying the BFB-n conditions and the conditions that point P (where P is either of \(A,B,C,C',D\)) is a local minimum. As mentioned above, any point satisfying these conditions with \(P=A\) is guaranteed to have the global minimum at A. In contrast, points obeying BFB-n and local minimum at \(P \in \{B,C,C',D\}\) can have the global minimum at any other point X, except \(X=A\). To be specific, let us generate points obeying BFB-n in Eq. (48) and the conditions for local minimum at \(P=C\) in Eq. (81). Then, we will find points where C is indeed the global minimum. But, we will also find points where \(X \in \{B,C',D\}\) is instead the global minimum. An example with \(P=C\) a local minimum and a global minimum at \(X=D\) below is given in Eq. (87).

Moreover, BFB-n conditions and the conditions that point P (where P is either of \(A,B,C,C',D\)) is a local minimum do not even preclude the existence of a charge breaking (CB) global minimum below. We found such examples when choosing \(P=D\) as a local minimum. One such example is

$$\begin{aligned}&\Lambda _0 = 2.92212\, , \ \ \ \Lambda _1 = -0.276035\, , \ \ \ \nonumber \\&\qquad \qquad \Lambda _2 = 2.58004\, , \nonumber \\&\Lambda _3 = 0.477722\, , \ \ \ \Lambda _4 = 1.80921\, , \end{aligned}$$
(111)

leading to

$$\begin{aligned}&V_A= -0.294131\, , \ \ \ V_B= -0.377917\, , \ \ \ \nonumber \\&\qquad \qquad V_C= -0.179483\, , \nonumber \\&V_C'= -0.249705\, , \ \ \ V_D= -0.379115\, , \end{aligned}$$
(112)

in units of \(M_0^2/4\). Thus, looking only at the neutral minima, D would indeed be the global minimum. However, in this case, there is a lower lying minimum at a CB point, yielding \(V_\text {CB}= -0.38331\,(M_0^2/4)\). In fact, the point in Eq. (111) survives even after imposing the global minimum conditions \(\Lambda _X - \Lambda _D >0\) for all neutral \(X \in \{ A,B,C,C'\}\).

Starting from (111), one can generate an even worse potential by gradually lowering \(\Lambda _0\) while keeping \(\Lambda _{1,2,3,4}\) fixed.Footnote 11 For a \(\Lambda _0\) (roughly) between 0.2844 and 0.313, all points pass the conditions for BFB-n and global-D in the neutral directions, but, nevertheless, correspond to potentials which are unbounded from below along the CB directions. For example, the potential corresponding to

$$\begin{aligned}&\Lambda _0 = 0.300000\, , \ \ \ \Lambda _1 = -0.276035\, , \ \ \ \nonumber \\&\qquad \qquad \Lambda _2 = 2.58004\, , \nonumber \\&\Lambda _3 = 0.477722\, , \ \ \ \Lambda _4 = 1.80921\, , \end{aligned}$$
(113)

obeys BFB-n and the requirement that D be the lowest lying neutral minimum but, still, is unbounded from below. This example disproves a conjecture in Ref. [15].Footnote 12

But we have found something quite interesting concerning BFB-c in our extensive simulation. We performed a simulation of over 10000 parameter points, followed by a numerical minimization searching for the global minimum. For each of these parameter points, the minimization was repeatedly attempted with 200 random seed points. Requiring only that some point P be a local minimum without requiring BFB-n did yield many cases where the potential was unbounded from below along the charge breaking directions. As (111) and (113) illustrate, we have also found some points which passed the stronger requirements of BFB-n with D the lowest lying neutral minimum, but which yielded potentials either with a lower-lying CB minimum or which were even unbounded from below along the CB directions.

However, in all cases we generated where BFB-n holds, \(P=A,B,C\), or \(C'\) is a local minimum, and where the potential lies lower than at any other neutral point \(X \ne P\), we found no lower lying CB minimum and no potential unbounded from below along CB directions. We propose this result here in the form of a Conjecture:

  • The combined requirements that BFB-n holds and that point P (where P is either of \(A,B,C,C'\)) is a global minimum guarantee that the potential is also BFB-c and also that there in no lower-lying CB global minimum.

The analytical proof of this conjecture remains as an open difficult problem.

16 Conclusions

We have revisited the conditions for global minimum in the exact \(A_4\) 3HDM, in light of the candidates for local minimum identified in [11] via the geometric minimization method. We stressed that C and \(C'\) are not equivalent, since \(V(C) \ne V(C')\). Given a set of \(\Lambda \) parameters, both must be probed as candidates for global minimum. For point A, we found that the conditions determined in [11] and repeated in [13] are indeed conditions that guarantee that A is the global minimum. In fact, we have found that, by themselves, they also imply that the potential is BFB-n. The conditions for local/global minimum at points B, C, \(C'\), and D are either absent or wrong in the literature [11, 13]. We present here the correct conditions for B, C, \(C'\), and D to be the global minimum.

The clarification provided here as to the number of global minima candidates and the exact conditions for each global minimum is also important given recent developments proposing specific types of soft-symmetry breaking which preserve de form (\(A,B,C,C'\) or D) of the critical points which are candidates for global minimum [13, 25]. In a numerical simulation, the correct conditions with an exact symmetry can be used to gain confidence in the results with soft-symmetry breaking, as the \(A_4\) soft-breaking terms are made smaller.

Finally, we address the issue of the charge breaking directions. We present a sufficient analytical condition for the potential to be BFB-c. We also perform an extensive numerical simulation, resulting in a conjecture about BFB-c, proposing that, after BFB-n and global-P, only points with \(P=D\) are at risk of allowing for a lower-lying CB (true) global minimum or unbounded from below potentials.