Quantum Flag Manifold $\sigma$-Models and Hermitian Ricci Flow

Abstract

We show that flag manifold $\sigma$-models (including $\mathbb{C}\mathbb{P}^{n-1}$, Grassmannian models as special cases) and their deformed versions may be cast in the form of gauged bosonic Thirring/Gross-Neveu-type systems. Quantum mechanically the gauging is violated by chiral anomalies, which may be cancelled by adding fermions. We conjecture that such models are integrable and check on some examples that the trigonometrically deformed geometries satisfy the generalized Ricci flow equations.

Appendices

Metric, B-Field and Dilaton of the Deformed $\mathbb {C}^n$-Model

Here we derive the formulas (3.24) from the main text. We start by writing out the quadratic form in the Lagrangian (3.1):

(A.1)

The constants $\upalpha , \upbeta , \upgamma$ were defined in (3.9). Inverting the quadratic form Q, we obtain the Lagrangian of the $\sigma$-model:

$$\begin{aligned} \mathscr {L}=\sum \limits _{k, m=1}^n\,\partial {\overline{U}}_k \left( \delta _{km}{{\hat{\lambda }}_k}-\frac{{\hat{\lambda }}_k {\hat{\lambda }}_m\,U_k {\overline{U}}_m}{\upkappa +\sum \limits _p\,{\hat{\lambda }}_p\,|U_p|^2}\right) {\overline{\partial }}U_m,\quad \upkappa =(b-{\upgamma \over n})^{-1}. \end{aligned}$$

(A.2)

It is convenient to introduce the parametrization $U_j=e^{t_j- i\,\phi _j}$ and the notation $\lambda _j:={\hat{\lambda }}_j \,e^{2t_j}$ (we already used it in Sect. 3.4). In this case $d U_j\wedge d{\overline{U}}_j=2i\,e^{2t_j}\,dt_j\wedge d\phi _j$ and $\sum \limits _{j=1}^n\,{\hat{\lambda }}_j\,U_j \,d{\overline{U}}_j=\sum \limits _{j=1}^n\,\lambda _j\,(dt_j- i\,d\phi _j)$. Therefore¹⁷

$$\begin{aligned}{} & {} ds^2= \sum \limits _{i, j=1}^n\,k_{ij} (dt_i\,dt_j+d\phi _i d\phi _j). \end{aligned}$$

(A.3)

$$\begin{aligned}{} & {} B= \sum \limits _{i, j=1}^n\,k_{ij}\, dt_i\wedge d\phi _j,\quad \text {where}\quad {1\over 2}\,k_{ij}=\lambda _i\,\delta _{ij}-\frac{\lambda _i \lambda _j}{\upkappa +\sum \limits _k \lambda _k}. \end{aligned}$$

(A.4)

The dilaton $\Phi$ arises from the integration over the V-variables, namely from the determinant of the quadratic form in (A.1):

$$\begin{aligned} e^{-2\Phi }=\textrm{Det}(Q)=\prod \limits _{k=1}^n\,{1\over {\hat{\lambda }}_k}\times \left( 1+{1\over \upkappa }\,\sum \limits _{k=1}^n{\hat{\lambda }}_k |U_k|^2\right) . \end{aligned}$$

(A.5)

1.1 $n=2$: the deformed $\mathbb {C}^2$

In computing the Ricci tensor and other geometric quantities it is much more convenient to deal with the inverse metric $\kappa =k^{-1}$, which has the simple form

(A.6)

After the shift of $t_i$-variables described in Sect. 3.2 the functions $\upchi _i$ are found to be

$$\begin{aligned} \upchi _1={1\over 2}\,{1+e^{2\tau }\over 1-e^{2\tau }}+{e^{\tau }\over 1-e^{2\tau }}\,e^{t_2-t_1},\quad \upchi _2={1\over 2}\,{1+e^{2\tau }\over 1-e^{2\tau }}+{e^{\tau }\over 1-e^{2\tau }}\,e^{t_1-t_2}. \end{aligned}$$

(A.7)

We start by computing the angular components $R^{\phi _i \phi _j}$ (again, with upper indices) of the Ricci tensor. These may be obtained using the formula $R^\mu _{\nu }\,K^\nu ={1\over \sqrt{g}}\,\partial _\rho \left( \sqrt{g}\,\nabla ^{\mu }K^\rho \right)$, valid for any Killing vector field K (cf. [82]). As it follows from (A.6), ${\partial \kappa ^{ij}\over \partial t_k}={1\over 2}{\partial \upchi _i\over \partial t_k}\,\delta ^{ij}$. Using this as well as the fact that $\upchi _i=\upchi _i(t_1-t_2)$, we arrive at the formula for the $\phi$-components of the Ricci tensor:

$$\begin{aligned} R^{\phi _i \phi _j}=\sum \limits _{p=1}^n\,{1\over 8k}\,{\partial \over \partial t_p}\left( k\,\upchi _p\,{\partial \upchi _i\over \partial t_p}\right) \,\delta _{ij}-k_{ij}\,\sum \limits _{p=1}^n\,{1\over 8}\,\upchi _p\,{\partial \upchi _i\over \partial t_p} \,{\partial \upchi _j\over \partial t_p},\quad k=\textrm{det}(k_{ij}).\nonumber \\ \end{aligned}$$

(A.8)

For other ingredients of the r.h.s. of (3.23) we also obtain expressions in terms of $\upchi _i$:

$$\begin{aligned} (H^2)^{\phi _i \phi _j}=-{1\over 2}{\partial \upchi _i\over \partial t_j}\,{\partial \upchi _j\over \partial t_i}+{1\over 2}\,k_{ij}\,\sum \limits _{p=1}^n\,\upchi _p\,{\partial \upchi _i\over \partial t_p} \,{\partial \upchi _j\over \partial t_p}\, \end{aligned}$$

(A.9)

$$\begin{aligned} 2\,\nabla ^{\phi _i} \nabla ^{\phi _j} \Phi ={1\over 8}\,\delta ^{ij}\,\sum \limits _{p=1}^n\,\upchi _p\,{\partial \upchi _i\over \partial t_p}\,\left( 2-{1\over k}\,{\partial k\over \partial t_p}\right) . \end{aligned}$$

(A.10)

Here we have used that the dilaton is $\Phi =-t_1-t_2+{1\over 2}\log {k}$. Assembling the pieces, we arrive at the following expression for the r.h.s of the Ricci flow Eq. (3.23):

(A.11)

Using the explicit expressions (A.7), we find that

$$\begin{aligned} {d \kappa ^{ij}\over d\tau }=R^{\phi _i \phi _j}+{1\over 4}\,(H^2)^{\phi _i \phi _j}+2\,\nabla ^{\phi _i} \nabla ^{\phi _j} \Phi . \end{aligned}$$

(A.12)

The mixed $(t, \phi )$-components of both sides of the Ricci flow equations are identically zero, so it remains to compute the (t, t)-components. Here we do not find any significant simplifications, so we just outline the strategy of how this can be done. The (t, t)-components of the Ricci tensor may be written as follows:

$$\begin{aligned} R_{t_i t_j}=R_{ij}^{(k)}-{1\over 2}\,\nabla _i \nabla _j \,\log {k}-{1\over 4}\,\textrm{Tr}\left( {\partial k^{-1}\over \partial t_i}\,k\, {\partial k^{-1}\over \partial t_j}\,k\right) \, \end{aligned}$$

(A.13)

where $R_{ij}^{(k)}$ is the Ricci tensor of the two-dimensional metric $k_{ij}$. As for any such metric, $R_{ij}^{(k)}={R\over 2}\,k_{ij}$, where R is the Ricci scalar. The latter can be computed by noticing that the metric has additional isometry $t_i\rightarrow t_i+\mathrm {const.}$ Introducing the variables $x=t_1-t_2$ and $y=t_1+t_2$ and shifting y to remove the cross-term, we bring the metric to diagonal form:

$$\begin{aligned} k_{ij}\,dt_i\,dt_j=\frac{2\,dx^2}{\upchi _1+\upchi _2}+{k\over 2}\,(\upchi _1+\upchi _2)\,dy^2,\quad k(x)={\upkappa \over \upchi _1+\upchi _2+\upkappa \,\upchi _1\upchi _2}. \end{aligned}$$

(A.14)

Denoting $\upchi _1+\upchi _2:=c(x)$, we calculate the Ricci scalar $2R=-c''-{k''\over k}\,c-{3\over 2}\,{k'\over k}\,c'+{c\over 2}\,{(k')^2\over k^2}$. One also has a simple formula for the Christoffel symbols with all upper indices: $\Gamma ^{p|mn}=-{1\over 8} \upchi _n\,\partial _n\upchi _m\,\delta ^{pm}-{1\over 8} \upchi _m\,\partial _m\upchi _n\,\delta ^{pn}+{1\over 8}\,\upchi _p\,\partial _p\upchi _n\,\delta ^{mn}$. The rest is a matter of direct calculation, which shows that the Ricci flow Eq. (3.23) is satisfied. It would be interesting to find a proper geometrical framework that would facilitate such calculations.

Computing the $\beta$-Function in the Inhomogeneous Gauge

In this appendix we consider the renormalization of the theory (4.14) at one loop. We wish to show that all divergences may be reabsorbed in a redefinition of the deformation parameter s entering $\upalpha , \upbeta , \upgamma$. The quartic vertices in (4.14) look exactly as in (2.9), but with the summation restricted to $1 \ldots n-1$ and the truncated r-matrix of the form

$$\begin{aligned} {\widetilde{r}}_{ij}^{kl}=a_{ij}\delta _i^k \delta _j^l+\upgamma \,\delta _{ij}\delta ^{kl},\quad i, j, k, l=1\ldots n-1. \end{aligned}$$

(B.1)

One contribution to the $\beta$-function comes from exactly the same diagrams as before, shown in Fig. 4, where in the vertices one replaces $r\rightarrow {\widetilde{r}}$:

$$\begin{aligned} \left( \beta _{ij}^{kl}\right) _{1}= & {} \left[ \frac{(n-1)s}{(1-s)^2}+(i-j)a_{ij}\right] \,\left( \delta _i^k \delta _j^l-{1\over n-1}\delta _{ij}\delta ^{kl}\right) \nonumber \\{} & {} =\left[ (n-1)\,\upalpha \upbeta +(i-j)a_{ij}\right] \,\delta _i^k \delta _j^l-\upalpha \upbeta \,\delta _{ij}\delta ^{kl}. \end{aligned}$$

(B.2)

Fig. 4

Diagrams contributing to $\left( \beta _{ij}^{kl}\right) _{1}$. Here $i, j, k, l=1, \ldots , n-1$

Fig. 5

Bubbles in the sextic vertex, contributing to $\left( \beta _{ij}^{kl}\right) _{2}$

An additional contribution comes from bubbles in the sextic vertex (see Fig. 5):

$$\begin{aligned} \left( \beta _{ij}^{kl}\right) _{2}=\upalpha \upbeta (n+1)\,\delta _{ij}\delta ^{kl}+\upalpha \upbeta \delta _i^k \delta _j^l. \end{aligned}$$

(B.3)

Summing the two contributions, we obtain

$$\begin{aligned} \beta _{ij}^{kl}=\left[ n\,\upalpha \upbeta +(i-j)a_{ij}\right] \,\delta _i^k \delta _j^l+n\upalpha \upbeta \,\delta _{ij}\delta ^{kl} \end{aligned}$$

(B.4)

As explained in Sect. 3.2, the part proportional to $(i-j)a_{ij}$ can be removed by a redefinition of coordinates, and the remaining terms give rise to the Ricci flow equation ${\dot{\upgamma }}=n\upalpha \upbeta$ that is solved by $s=e^{n\tau }$.

Fig. 6

A bubble in the quartic vertex, contributing to $\upbeta \,|V_k|^2$ terms

As another illustration let us consider the renormalization of the term $\upbeta \,|V_k|^2$ in (4.14). The divergent graphs that contribute to the $\beta$-function of this ‘vertex’ come from bubbles in the quartic vertex, see Fig. 6:

$$\begin{aligned} \beta _{i}^j=\delta _{i}^j\,\upbeta \,\sum \limits _{k=1}^{n-1}\,{\widetilde{r}}_{\,ki}^{\,kj}=\delta _{i}^j\,\upbeta \,\left( \sum \limits _{k=1}^{n-1}\,a_{kj}+\upgamma \right) =\delta _{i}^j\,\upbeta \,((n-j)+\upalpha \, n) \end{aligned}$$

(B.5)

The first term in brackets is again cancelled by a redefinition of variables. We showed in Sect. 3.2 that the U-variables evolve with the Ricci flow as $|U_j|^2=e^{j\,\tau }$. However, since we have chosen a gauge $U_n=1$, we have to perform a compensating gauge transformation, and as a result the evolution of the variables is given by $|U_j|^2=e^{(j-n)\,\tau }$. The V-variables evolve as inverses of these, $|V_j|^2=e^{-(j-n)\,\tau }$, and this eliminates the unwanted term in (B.5). The remaining flow equation is ${\dot{\upbeta }}=n\upalpha \upbeta$, which is again solved by $s=e^{n\tau }$. A similar analysis could be performed for other vertices in the Lagrangian (4.14).

The Deformed Flag Manifold $\sigma$-Model in Geometric Form

We start this appendix by showing that there is a GLSM formulation of the flag manifold $\sigma$-model based on $n\times n$-matrices rather than on $m\times n$ matrices as in Sect. 4.2. The quotient (4.20) suggests the following GLSM representation for a deformed flag manifold $\sigma$-model:

$$\begin{aligned} \mathscr {L}=\textrm{Tr}\left( V {\overline{D}} U\right) -\textrm{Tr}\left( V {\overline{D}} U\right) ^\dagger +\textrm{Tr}\left( r_s^{\textrm{T}}(U V) (U V)^\dagger \right) , \end{aligned}$$

(C.1)

where, as opposed to (4.21), $U, V\in \textrm{End}(\mathbb {C}^n)$ are square matrices. The covariant derivatives are defined formally as before: ${\overline{D}}U={\overline{\partial }}U+i \,U \overline{{\mathcal {A}}}$, but now one should keep in mind that the gauge field takes values in $\mathfrak {p}=\textrm{Lie}(P_{m_1, \ldots , m_{s}})\subset {\mathfrak {g}}{\mathfrak {l}}_n$. As we now explain, one can simplify this gauged theory to arrive at the Lagrangian (4.21) studied before, if one assumes that U is non-degenerate, i.e. $U\in GL(n, \mathbb {C})$. Taking the variation of (C.1) w.r.t. $\overline{{\mathcal {A}}}$, we find $VU|_{\mathfrak {p}^\vee }=0$ (here we view $VU\in {\mathfrak {g}}{\mathfrak {l}}_n^\vee$). In general this is a rather non-trivial condition, however already from the requirement of vanishing of the first $n_s$ rows of the matrix VU we find that the first $n_s$ rows of V are orthogonal to all columns of U. Since U is non-generate, it follows that $V_{ik}=0$ for $i=1, \ldots , n_s$. Therefore we can safely truncate the matrix V by erasing its first $n_s$ rows, forming a reduced matrix ${\widetilde{V}}\in \textrm{Hom}(\mathbb {C}^n, \mathbb {C}^m)$ with $m=\sum \limits _{i=1}^{s-1}\,n_i$. It is then easy to see that the first $n_s$ columns of the matrix U do not enter the Lagrangian (C.1) either, so we may analogously truncate the matrix U, forming a new matrix ${\widetilde{U}}\in \textrm{Hom}(\mathbb {C}^m, \mathbb {C}^n)$. We may then replace the fields U and V in the Lagrangian (C.1) by ${\widetilde{U}}, {\widetilde{V}}$ and truncate the gauge fields in the covariant derivatives accordingly. If one drops the tildes, one recognizes the formula (4.21).

Our next goal is to derive the metric form of the deformed flag manifold model (4.21), where U and V are $m\times n$-matrices. If necessary, one can always return from (4.21) to (C.1) by setting $m=n$, so in a sense the $m\ne n$ notation is more general. Our goal in this section is to eliminate the variables $V, {\overline{V}}$, which enter the Lagrangian quadratically. The relevant equations to be solved are (2.12)–(2.13) with ${\hat{r}}=r$. To write down the solution, we introduce the projector $\Pi _U=U({\overline{U}} U)^{-1}{\overline{U}}$ on $\textrm{Im}(U)$ (indeed $\Pi _U(UY)=UY$). Then

$$\begin{aligned} V={1\over {\overline{U}} U} \,{\overline{U}} \left[ \frac{1}{\frac{1}{2}\,\frac{1+s}{1-s}\,\textrm{Id}+\Pi _U\,{i\over 2}\,{\mathcal {R}}}\right] \, \left( U {1\over {\overline{U}} U} D{\overline{U}} \right) . \end{aligned}$$

(C.2)

Using $\Pi _U U=U$ and $\overline{U} \Pi _U= \overline{U}$, it is easy to prove that in the denominator one can freely commute $\Pi _U$ with ${\mathcal {R}}$. One shows that (2.13) is satisfied by noting that ${\overline{U}}\,r_s\,\Pi _U={\overline{U}} \left( \frac{1}{2}\,\frac{1+s}{1-s}\,\textrm{Id}+{i\over 2}\,{\mathcal {R}} \,\Pi _U\, \right)$ and using this commutation property.

The formula (C.2) for V can be considerably simplified by using part of the complex gauge symmetry. We will call this procedure ‘passing to the unitary frame’, for the following reason. We may write $U={\hat{U}}{\hat{b}}$, where $\overline{{\hat{U}}} {\hat{U}}=\mathbb {1}_m$ and ${\hat{b}}\in B\subset P_{m_1, \ldots , m_s}$ is a lower-triangular matrix (B is the Borel subgroup). This decomposition is simply the Gram–Schmidt orthogonalization of the m vectors in U. The remaining gauge group is then $H=U(n_1)\times \cdots \times U(n_s)$, which is of course the denominator of the quotient (4.17). The solution for V clearly simplifies in the unitary gauge ${\overline{U}} U=\mathbb {1}_m$. Using this gauge and inserting V back into the Lagrangian, we find

$$\begin{aligned} \mathscr {L}\sim \textrm{Tr}\left( \left( U D{\overline{U}} \right) ^\dagger \left[ \frac{1}{\frac{1}{2}\,\frac{1+s}{1-s}\,\textrm{Id}+U{\overline{U}}\,{i\over 2}\,{\mathcal {R}}}\right] \, \left( U D{\overline{U}} \right) \right) \end{aligned}$$

(C.3)

An additional simplification occurs, if we use the formulation of the model where U is an $n\times n$-matrix. In that case the unitary gauge implies that U is unitary, so that, apart from ${\overline{U}} U=\mathbb {1}_n$ we have $U {\overline{U}}=\mathbb {1}_n$, and as a result one can forget the projector in the denominator in (C.3), arriving at the simple formula

Of course in this case the complications are absorbed in the enlarged gauge field ${\mathcal {A}}$.