Smooth Maps Minimizing the Energy and the Calibrated Geometry

Abstract

We generalize the notion of calibrated submanifolds to smooth maps and show that several kinds of smooth maps appearing in the differential geometry are applicable to our situation. Moreover, we apply this notion to give the lower bound to some energy functionals of smooth maps in the given homotopy class between Riemannian manifolds and consider the energy functional which is minimized by the identity maps on the Riemannian manifolds with special holonomy groups.

1 Introduction

In this article, we introduce the notion of calibrated geometry for smooth maps between Riemannian manifolds and consider the lower bound or the minimizers of several energy of smooth maps. Let (X, g) and (Y, h) be compact Riemannian manifolds and $f:X\rightarrow Y$ be a smooth map. Then the p-energy of f is defined by

$$\begin{aligned} \mathcal {E}_p(f):=\int _X |\textrm{d}f|^p\textrm{d}\mu _g \end{aligned}$$

for $p\ge 1$, where $\mu _g$ is the volume measure of g. A harmonic map is a critical point of $\mathcal {E}_2$ and it is studied well by many researchers in differential geometry. In 1964, Eells and Sampson [4] have shown that there is a harmonic map $f'$ homotopic to f if the sectional curvature of h is nonpositive. Moreover, Hartman [5] showed that such harmonic maps minimize $\mathcal {E}_2|_{[f]}$, where [f] is the homotopy class represented by f.

In general, harmonic maps need not minimize the energy. For example, although the identity maps on any Riemannian manifolds are always harmonic, it is known that there is a family of smooth maps $\{ f_\varepsilon \}_{\varepsilon >0}$ homotopic to the identity map of the n-sphere $S^n$ with the standard metric such that $\lim _{\varepsilon \rightarrow 0}\mathcal {E}_2(f_\varepsilon )=0$, if $n\ge 3$. By the result shown by White [10], if $\pi _l(X)$ is trivial for all $1\le l\le k$, then $\inf \mathcal {E}_k|_{[1_X]}=0$, where $1_X$ is the identity map of X.

One of the motivations of this article is to give the lower bound to the energy restricting to a given homotopy class [f] and the minimizer of them. Such a lower bound was first obtained by Lichnerowicz [8] in the case of (X, g) and (Y, h) are Kähler manifolds, then it was shown that any holomorphic maps between Kähler manifolds minimize $\mathcal {E}_2$ in their homotopy classes. Moreover, Croke [3] showed that the identity map on the real projective space with the standard metric minimize $\mathcal {E}_2$ in its homotopy class, then Croke and Fathi [2] introduced the new homotopy invariant called the intersection, which gives the lower bound to $\mathcal {E}_2|_{[f]}$ for a given homotopy class [f]. Recently, Hoisington [7] give the lower bound to $\mathcal {E}_p$ for an appropriate p in the case of X is real, complex, or quaternionic projective spaces with the standard metrics.

In this article, we generalize the notion of calibrated geometry to smooth maps between smooth manifolds, which give the lower bound to several energies. The origin of calibrated geometry is the Wirtinger’s inequality for the even-dimensional subspaces in Hermitian inner product spaces [11], then it refined or generalized by many researchers. In [6], Harvey and Lawson defined calibrated submanifolds in the Calabi–Yau, $G_2$ or $\textrm{Spin}(7)$ manifolds which minimize the volume in their homology classes. Similarly, we define the new class of smooth maps, called calibrated maps, and show that they minimize the appropriate energy for the given situation. Moreover, we obtain the next results as applications.

The first application is to obtain the lower bound to p-energy restricting to the given homotopy class. We assume X is oriented. The pullback of f induces a linear map $[f^*]^k:H^k(Y,{\mathbb {R}})\rightarrow H^k(X,{\mathbb {R}})$. By fixing basis of $H^k(X,{\mathbb {R}})$, $H^k(Y,{\mathbb {R}})$, we obtain the matrix $P([f^*]^k)$ of $[f^*]^k$ and put $|P([f^*]^k)|:=\sqrt{\textrm{tr}({}^tP([f^*]^k)\cdot P([f^*]^k))}$.

Theorem 1.1

Let (X, g) and (Y, h) be as above. For any $1\le k\le \dim X$, there is a positive constant C depending only on k, (X, g), (Y, h) and the basis of $H^k(X,{\mathbb {R}})$, $H^k(Y,{\mathbb {R}})$ such that for any $f\in C^\infty (X,Y)$, we have

$$\begin{aligned} \mathcal {E}_k(f)\ge C|P([f^*]^k)|. \end{aligned}$$

In particular, if $[f^*]^k$ is nonzero, then $\inf (\mathcal {E}_k|_{[f]})$ is positive.

In the above theorem, the compactness of Y is not essential. See Theorem 4.2.

The second application is to show that the identity maps of some Riemannian manifolds with special holonomy groups minimize the appropriate energy. As we have already mentioned, the identity map on the real or complex projective space minimizes $\mathcal {E}_2$ in its homotopy class by [3] and [8], respectively. It was shown by Wei [9] that the identity map on the quaternionic projective space ${\mathbb {H}}{\mathbb {P}}^n$ with the standard metric is an unstable critical point of $\mathcal {E}_p$ for $1\le p< 2+4n/(n+1)$. Moreover, Hoisington gave the nontrivial lower bound of $\mathcal {E}_p|_{[1_{{\mathbb {H}}{\mathbb {P}}^n}]}$ for $p\ge 4$. Here, the quaternionic projective space is a typical example of quaternionic Kähler manifolds, which are Riemannian manifolds of dimension 4n whose holonomy group is contained in $Sp(n)\cdot Sp(1)$. Now, let A be an $n\times m$ real-valued matrix and denote by $a_1,\ldots ,a_m\in {\mathbb {R}}$ the nonnegative eigenvalues of ${}^tAA$, then put $|A|_p:=(\sum _{i=1}^m a_i^{p/2})^{1/p}$. Moreover, we define an energy $\mathcal {E}_{p,q}$ by

$$\begin{aligned} \mathcal {E}_{p,q}(f):=\int _X|\textrm{d}f|_p^q\textrm{d}\mu _g, \end{aligned}$$

then we have $\mathcal {E}_p=\mathcal {E}_{2,p}$.

Theorem 1.2

Let (X, g) be a compact quaternionic Kähler manifold of dimension $4n\ge 8$. Then the identity map of X minimizes $\mathcal {E}_{4,4}$ in its homotopy class.

We can also show the similar theorem in the case of other holonomy groups. If (X, g) is a compact $G_2$ manifold, then $1_X$ minimizes $\mathcal {E}_{3,3}|_{[1_X]}$ and if (X, g) is a compact $\textrm{Spin}(7)$ manifold, then $1_X$ minimizes $\mathcal {E}_{4,4}|_{[1_X]}$ (see Theorem 5.6). Moreover, it is easy to see that if the identity map minimizes $\mathcal {E}_{p,q}$, then it also minimizes $\mathcal {E}_{p',q'}$ for all $p'\ge p$ and $q'\ge q$ by the Hölder’s inequality. Of course, we can also consider the case of Kähler, Calabi–Yau, and hyper-Kähler manifolds, respectively; however, the results in these cases also follow from [8].

This paper is organized as follows. In Sect. 2, we define the notion of calibrated maps, which is the analogy of the calibrated submanifolds. In Sect. 3, we explain some examples of calibrated maps. We show that holomorphic maps between Kähler manifolds and the inclusion maps of calibrated submanifolds can be regarded as calibrated maps. Moreover, we can also show that the fibration whose regular fibers are calibrated submanifolds are calibrated maps. We prove Theorem 1.1 in Sect. 4, and Theorem 1.2 in Sect. 5. In Sect. 6, we compare the homotopy invariant introduced in [2] with the invariants defined in this paper.

2 Calibrated Maps

Let X, Y be smooth manifold of $\dim X=m$ and $\dim Y=n$. Throughout of this paper, we suppose X is compact and oriented. We fix a volume form $\textrm{vol}\in \Omega ^m(X)$ on X, namely, a nowhere vanishing m-form which determines an orientation and a measure of X. For m-forms $v_1,v_2\in \Omega ^m(X)$, there are $\varphi _i\subset C^\infty (X)$ with $v_i=\varphi _i\textrm{vol}$. Then we write $v_1\le v_2$ if $\varphi _1(x)\le \varphi _2(x)$ for all $x\in X$.

If a map $\sigma :C^\infty (X,Y)\rightarrow L^1(X)$ is given, then we can define an energy $\mathcal {E}:C^\infty (X,Y)\rightarrow {\mathbb {R}}$ by

$$\begin{aligned} \mathcal {E}(f):=\int _X \sigma (f)\textrm{vol}. \end{aligned}$$

Now, $f_0,f_1\in C^\infty (X,Y)$ are said to be homotopic if there is a smooth map $F:[0,1]\times X\rightarrow Y$ such that $F(0,\cdot )=f_0$ and $F(1,\cdot )=f_1$. By Whitney approximation theorem, it is equivalent to the existence of the continuous homotopy joining $f_0$ and $f_1$. For $f\in C^\infty (X,Y)$, denote by $[f]\subset C^\infty (X,Y)$ the homotopy equivalent class represented by f. In this paper, we consider the lower bound to $\mathcal {E}|_{[f]}$ or the minimum of $\mathcal {E}|_{[f]}$.

Denote by $1_X:X\rightarrow X$, the identity map on X. We define a smooth map $(1_X,f):X\rightarrow X\times Y$ by

$$\begin{aligned} (1_X,f)(x):=(x,f(x)). \end{aligned}$$

The next definition is the analogy of [6].

Definition 2.1

$\Phi \in \Omega ^m(X\times Y)$ is a $\sigma$-calibration if $\textrm{d}\Phi =0$ and

$$\begin{aligned} (1_X,f)^*\Phi \le \sigma (f)\textrm{vol}\end{aligned}$$

for any smooth map $f:X\rightarrow Y$. Moreover, f is a $(\sigma ,\Phi )$-calibrated map if

$$\begin{aligned} (1_X,f)^*\Phi = \sigma (f)\textrm{vol}. \end{aligned}$$

Theorem 2.2

Let $\sigma$ be an energy density and $\Phi$ be a $\sigma$-calibration.

• (i) The constant $\int _X(1_X,f)^*\Phi$ is determined by the homotopy class [f]. In other words, $\int _X(1_X,f_0)^*\Phi =\int _X(1_X,f_1)^*\Phi$ if $[f_0]=[f_1]$.

• (ii) We have $\inf \mathcal {E}|_{[f]} \ge \int _X(1_X,f)^*\Phi$ for any $f\in C^\infty (X,Y)$.

• (iii) We have $\mathcal {E}(f)=\int _X(1_X,f)^*\Phi$ iff f is $(\sigma ,\Phi )$-calibrated map. In particular, any $(\sigma ,\Phi )$-calibrated map minimizes $\mathcal {E}$ in its homotopy class.

Proof

$(\textrm{i})$ If $f_0,f_1$ are homotopic, then $(1_X,f_0)$ and $(1_X,f_1)$ are homotopic, accordingly $(1_X,f_0)^*\Phi$ and $(1_X,f_1)^*\Phi$ represent the same cohomology class by [1, Corollary 4.1.2].

$(\textrm{ii})$ follows from the definition of $\sigma$-calibration.

$(\textrm{iii})$ By the point-wise inequality $(1_X,f)^*\Phi \le \sigma (f)\textrm{vol}$, we have $\mathcal {E}(f)=\int _X(1_X,f)^*\Phi$ iff $(1_X,f)^*\Phi = \sigma (f)\textrm{vol}$. $\square$

3 Examples

One of the typical example of the energy of maps is p-energy defined for the smooth maps between Riemannian manifolds. Let (X, g) and (Y, h) be Riemannian manifolds and $f:X\rightarrow Y$ be a smooth map. Then the pullback $f^*h$ is a section of $T^*X\otimes T^*X$, and we can take the trace $\textrm{tr}_g(f^*h)$. For $p\ge 1$, put $\sigma _p(f):=\{ \textrm{tr}_g(f^*h)\}^{p/2}$. We assume that X is oriented and denoted by $\textrm{vol}_g$ the volume form of g. The p-energy $\mathcal {E}_p(f)$ is defined by

$$\begin{aligned} \mathcal {E}_p(f):=\int _X\sigma _p(f)\textrm{vol}_g. \end{aligned}$$

Now, the differential $\textrm{d}f_x$ is an element of $T^*_xX\otimes T_{f(x)}Y$ for every $x\in X$. Since $g_x$ and $h_{f(x)}$ induce the natural inner product and the norm on $T^*_xX\otimes T_{f(x)}Y$, then we may also write $\sigma _p(f)(x)=|\textrm{d}f_x|^p$.

By the Hölder’s inequality, we have

$$\begin{aligned} \mathcal {E}_p(f)&\le \textrm{vol}_g(X)^{1-p/q}\mathcal {E}_q(f)^{p/q} \end{aligned}$$

for $1\le p\le q$. Thus, we have the following proposition.

Proposition 3.1

Let $\Phi \in \Omega ^m(X\times Y)$ be a $\sigma _p$-calibration. Then

$$\begin{aligned} \textrm{vol}_g(X)^{-1+p/q}\int _X(1_X,f)^*\Phi \le \mathcal {E}_q(f)^{p/q} \end{aligned}$$

for any $q\ge p$ and $f\in C^\infty (X,Y)$.

3.1 Holomorphic Maps

Here, assume that X, Y are complex manifolds and g, h are Kähler metrics. Let $m=\dim _{\mathbb {C}}X$ and $n=\dim _{\mathbb {C}}Y$. Then we have the decomposition

$$\begin{aligned} T^*X\otimes {\mathbb {C}}&=\Lambda ^{1,0}T^*X\oplus \Lambda ^{0,1}T^*X,\\ TY\otimes {\mathbb {C}}&=T^{1,0}Y\oplus T^{0,1}Y, \end{aligned}$$

accordingly the derivative $\textrm{d}f\in \Gamma (T^*X\otimes f^*TY)$ is decomposed into

$$\begin{aligned} \textrm{d}f&=(\partial f)^{1,0}+(\partial f)^{0,1}+(\overline{\partial }f)^{1,0}+(\overline{\partial }f)^{0,1}\\&\in (\Lambda ^{1,0}T^*X\otimes T^{1,0}Y) \oplus (\Lambda ^{1,0}T^*X\otimes T^{0,1}Y)\\&\quad \oplus (\Lambda ^{0,1}T^*X\otimes T^{1,0}Y) \oplus (\Lambda ^{0,1}T^*X\otimes T^{0,1}Y). \end{aligned}$$

Since $\textrm{d}f$ is real, we have

$$\begin{aligned} \overline{(\partial f)^{1,0}}=(\overline{\partial }f)^{0,1}, \quad \overline{(\partial f)^{0,1}}=(\overline{\partial }f)^{1,0}. \end{aligned}$$

Denote by $\omega _g,\omega _h$ the Kähler form of g, h, respectively, then the volume form is given by $\textrm{vol}_g=\frac{1}{m!}\omega _g^m$. The following observation was given by Lichnerowicz.

Theorem 3.2

[8] For any smooth map $f:X\rightarrow Y$, we have

$$\begin{aligned} \omega _g^{m-1}\wedge f^*\omega _h&=(m-1)!(|(\partial f)^{1,0}|^2-|(\overline{\partial }f)^{1,0}|^2)\textrm{vol}_g,\\ |\textrm{d}f|^2&=2|(\partial f)^{1,0}|^2+2|(\overline{\partial }f)^{1,0}|^2. \end{aligned}$$

In particular, we have

$$\begin{aligned} \mathcal {E}_2(f)\ge \frac{2}{(m-1)!}\int _X\omega _g^{m-1}\wedge f^*\omega _h \end{aligned}$$

and the equality holds iff f is holomorphic.

Now, we consider $\omega _g^{m-1}\wedge \omega _h\in \Omega ^m(X\times Y)$. The first two equalities in Theorem 3.2 implies that $\frac{2}{(m-1)!}\omega _g^{m-1}\wedge \omega _h$ is a $\sigma _2$-calibration. Moreover, the second statement implies that f is a $(\sigma _2,\frac{2}{(m-1)!}\omega _g^{m-1}\wedge \omega _h)$-calibrated map iff f is holomorphic. One can also see that f is $(\sigma _2,-\frac{2}{(m-1)!}\omega _g^{m-1}\wedge \omega _h)$-calibrated map iff f is anti-holomorphic.

3.2 Calibrated Submanifolds

In this subsection, we see the relation between the calibrated submanifolds in the sense of [6] and the calibrated maps. We assume $(Y^n,h)$ is a Riemannian manifold.

Definition 3.3

[6] For an integer $0<m<n$, $\psi \in \Omega ^m(Y)$ is a calibration if $\textrm{d}\psi =0$ and

$$\begin{aligned} \psi |_V\le \textrm{vol}_{h|_V} \end{aligned}$$

for any $y\in Y$ and m-dimensional oriented subspace $V\subset T_yY$. Here, $h|_V$ is the induced metric on V and $\textrm{vol}_{h|_V}$ is its volume form whose orientation is compatible with the one equipped with V. Moreover, an oriented submanifold $X\subset Y$ is a calibrated submanifold if

$$\begin{aligned} \psi |_{T_xX}=\textrm{vol}|_{h|_{T_xX}} \end{aligned}$$

for any $x\in X$.

Now, if X is an oriented manifold with a volume form $\textrm{vol}\in \Omega ^m(X)$, then for every linear map, $A:T_xX\rightarrow T_yY$ can be regarded as an $n\times m$-matrix by taking a basis $e_1,\ldots ,e_m$ of $T_xX$ and an orthonormal basis of $T_yY$ with $\textrm{vol}_x(e_1,\ldots ,e_m)=1$. Then $\sqrt{\det ( {}^tA\cdot A)}$ does not depend on the choice of these basis. Therefore, for $f\in C^\infty (X,Y)$, we can define the energy density $\tau _m(f)(x):=\sqrt{\det ({}^t\textrm{d}f_x\cdot \textrm{d}f_x)}$ and the energy $\mathcal {E}_{\tau _m}(f):=\int _X\tau _m(f)\textrm{vol}$.

Proposition 3.4

Let $(X,\textrm{vol})$ be an oriented manifold equipped with a volume form and $\psi \in \Omega ^m(Y)$ be closed. Assume that $\dim _{\mathbb {R}}X=m<n$ and denote by $\pi _Y:X\times Y\rightarrow Y$ the natural projection. Then $\psi$ is a calibration iff $\pi _Y^*\psi \in \Omega ^m(X\times Y)$ is a $\tau _m$-calibration. Moreover, for any embedding $f:X\rightarrow Y$, the following conditions are equivalent.

• (i) f(X) is a calibrated submanifold, where the orientation of f(X) is determined such that f preserves the orientation.

• (ii) f is a $(\tau _m,\pi _Y^*\psi )$-calibrated map.

Proof

Note that $(1_X,f)^*(\pi _Y^*\psi )=f^*\psi$ and $\tau _m(f)\textrm{vol}_g=\textrm{vol}_{f^*h}$. Hence $\psi$ is a calibration iff $\pi _Y^*\psi \in \Omega ^m(X\times Y)$ is a $\tau _m$-calibration. Moreover, suppose that f is an embedding. Then f is a $(\tau _m,\pi _Y^*\psi )$-calibrated map iff f(X) is a calibrated submanifold. $\square$

3.3 Fibrations

Let $(X^m,g)$ be an oriented Riemannian manifold and $Y^n$ be a smooth manifold equipped with a volume form $\textrm{vol}_Y\in \Omega ^n(Y)$. Here, we suppose $n< m$ and let $\varphi \in \Omega ^{m-n}$ be a calibration in the sense of Definition 3.3. Fix an orthonormal basis of $T_xX$ and a basis $e_1',\ldots ,e_n'\in T_yY$ with $\textrm{vol}_Y(e_1',\ldots ,e_n')=1$, we can regard a linear map $A:T_xX\rightarrow T_yY$ as an $m\times n$-matrix. Then the value of $\sqrt{\det (A\cdot {}^tA)}$ does not depend on the choice of above basis. For a smooth map $f:X\rightarrow Y$, put $\tilde{\tau }_{m,n}(f)|_x:=\sqrt{\det (\textrm{d}f_x\cdot {}^t\textrm{d}f_x)}$ and $\Phi :=\textrm{vol}_Y\wedge \varphi$.

Put

$$\begin{aligned} X_{\textrm{reg}}:=\{ x\in X|\, x \text{ is } \text{ a } \text{ regular } \text{ point } \text{ of } f\}. \end{aligned}$$

Note that $X_{\textrm{reg}}$ is open in X. If $x\in X_{\textrm{reg}}$, we have the orthogonal decomposition $T_xX=\textrm{Ker}(\textrm{d}f_x)\oplus H$ and $\textrm{d}f_x|_H:H\rightarrow T_{f(x)}Y$ is a linear isomorphism. Put $y=f(x)$ and suppose that $f^{-1}(y)$ is a calibrated submanifold with respect to the suitable orientation. We say that $\textrm{d}f_x$ is orientation preserving if there is a basis $v_1,\ldots ,v_m$ of $T_xX$ such that

$$\begin{aligned}&v_1,\ldots ,v_n \in H, \quad \textrm{vol}_Y(\textrm{d}f_x(v_1),\ldots ,\textrm{d}f_x(v_n))>0,\\&v_{n+1},\ldots ,v_m \in \textrm{Ker}(\textrm{d}f_x),\quad \varphi _x(v_{n+1},\ldots ,v_m)>0,\\&\textrm{vol}_g(v_1,\ldots ,v_m) >0. \end{aligned}$$

Proposition 3.5

$\Phi$ is a $\tilde{\tau }_{m,n}$-calibration. Moreover, a smooth map $f:X\rightarrow Y$ is a $(\tilde{\tau }_{m,n},\Phi )$-calibrated map iff

• (i) $f^{-1}(y)\cap X_{\textrm{reg}}$ is a calibrated submanifold with respect to $\varphi$ and the suitable orientation for any $y\in Y$,

• (ii) $\textrm{d}f_x$ is orientation preserving for any $x\in X_{\textrm{reg}}$.

Proof

If $x\in X$ is a critical point of f, then we can see

$$\begin{aligned} f^*\textrm{vol}_Y\wedge \varphi |_x=\tilde{\tau }_{m,n}(f)\textrm{vol}_g|_x=0. \end{aligned}$$

Fix a regular point x and an oriented orthonormal basis $e_1,\ldots ,e_m\in T_xX$ such that $e_{m-n+1},\ldots ,e_m\in \textrm{Ker}(\textrm{d}f_x)$. Then we have

$$\begin{aligned} f^*\textrm{vol}_Y\wedge \varphi (e_1,\ldots ,e_m)&=\textrm{vol}_Y(\textrm{d}f_x(e_1),\ldots ,\textrm{d}f_x(e_n)) \varphi (e_{n+1},\ldots ,e_m),\\ \tilde{\tau }_{m,n}(f)|_x&= \left| \textrm{vol}_Y(\textrm{d}f_x(e_1),\ldots ,\textrm{d}f_x(e_n))\right| . \end{aligned}$$

Since $\varphi$ is a calibration, we have $\varphi (\pm e_{n+1},e_{n+2},\ldots ,e_m)\le 1$, hence $|\varphi (e_{n+1},\ldots ,e_m)|\le 1$. Therefore,

$$\begin{aligned} f^*\textrm{vol}_Y\wedge \varphi (e_1,\ldots ,e_m)&\le \left| \textrm{vol}_Y(\textrm{d}f_x(e_1),\ldots ,\textrm{d}f_x(e_{m-n}))\right| = \tilde{\tau }_{m,n}(f)|_x, \end{aligned}$$

which implies that $\Phi$ is a $\tilde{\tau }_{m,n}$-calibration.

Next we consider the condition

$$\begin{aligned} f^*\textrm{vol}_Y\wedge \varphi |_x=\tilde{\tau }_{m,n}(f)\textrm{vol}_g|_x, \end{aligned}$$

where x is a regular value of f. In this case, we have the orthogonal decomposition $T_xX=\textrm{Ker}(\textrm{d}f_x)\oplus H$, where H is an n-dimensional subspace. We can take an orthonormal basis $e_1,\ldots ,e_m\in T_xX$ such that

$$\begin{aligned}&e_1,\ldots ,e_n \in H,\\&e_{n+1},\ldots ,e_m \in \textrm{Ker}(\textrm{d}f_x),\\&a:=\textrm{vol}_Y(\textrm{d}f_x(e_1),\ldots ,\textrm{d}f_x(e_{m-n}))>0,\\&\textrm{vol}_g(e_1,\ldots ,e_m) >0. \end{aligned}$$

Then we have

$$\begin{aligned} f^*\textrm{vol}_Y\wedge \varphi (e_1,\ldots ,e_m)&=a \varphi (e_{n+1},\ldots ,e_m),\\ \tilde{\tau }_{m,n}(f)\textrm{vol}_g(e_1,\ldots ,e_m)&=|a|=a. \end{aligned}$$

Therefore, we have

$$\begin{aligned} f^*\textrm{vol}_Y\wedge \varphi |_x&=\tilde{\tau }_{m,n}(f)\textrm{vol}_g|_x \end{aligned}$$

iff $\varphi (e_{n+1},\ldots ,e_m)=1$. Now we have taken $x\in X_{\textrm{reg}}$ arbitrarily, hence we have

$$\begin{aligned} f^*\textrm{vol}_Y\wedge \varphi&=\tilde{\tau }_{m,n}(f)\textrm{vol}_g \end{aligned}$$

iff $f^{-1}(y)\cap X_{\textrm{reg}}$ is a calibrated submanifold for any $y\in Y$ and $\textrm{d}f_x$ is orientation preserving for all $x\in X_{\textrm{reg}}$. $\square$

3.4 Totally Geodesic Maps Between Tori

Let ${\mathbb {T}}^n={\mathbb {R}}^n/{\mathbb {Z}}^n$ be the n-dimensional torus and we consider smooth maps from ${\mathbb {T}}^m$ to ${\mathbb {T}}^n$. Let $G=(g_{ij})\in M_m({\mathbb {R}})$ and $H=(h_{ij})\in M_n({\mathbb {R}})$ be positive symmetric matrices. Denote by $x=(x^1,\ldots ,x^m)$ and $y=(y^1,\ldots ,y^n)$ the Cartesian coordinate on ${\mathbb {R}}^m$ and ${\mathbb {R}}^n$, respectively, then we have closed 1-forms $\textrm{d}x^i\in \Omega ^1({\mathbb {T}}^m)$ and $\textrm{d}y^i\in \Omega ^1({\mathbb {T}}^n)$. We define the flat Riemannian metrics $g=\sum _{i,j}g_{ij}\textrm{d}x^i\otimes \textrm{d}x^j$ on ${\mathbb {T}}^m$ and $h=\sum _{i,j}h_{ij}\textrm{d}y^i\otimes \textrm{d}y^j$ on ${\mathbb {T}}^n$.

For a smooth map $f:{\mathbb {T}}^m\rightarrow {\mathbb {T}}^n$, we have the pullback $f^*:H^1({\mathbb {T}}^n,{\mathbb {R}})\rightarrow H^1({\mathbb {T}}^m,{\mathbb {R}})$. Here, since

$$\begin{aligned} H^1({\mathbb {T}}^m,{\mathbb {Z}})&=\textrm{span}_{\mathbb {Z}}\{ [\textrm{d}x^1],\ldots ,[\textrm{d}x^m]\},\\ H^1({\mathbb {T}}^n,{\mathbb {Z}})&=\textrm{span}_{\mathbb {Z}}\{ [\textrm{d}y^1],\ldots ,[\textrm{d}y^n]\}, \end{aligned}$$

there is $P=(P_i^j)\in M_{m,n}({\mathbb {Z}})$ such that $f^*[\textrm{d}y^j]=\sum _i P_i^j[\textrm{d}x^i]$. The matrix P is determined by the homotopy class of f. Now, let $*_g$ be the Hodge star operator of g and put

$$\begin{aligned} \Phi :=\sum _{i,j,k}h_{jk} P_i^j *_g\textrm{d}x^i\wedge \textrm{d}y^k \in \Omega ^m({\mathbb {T}}^m\times {\mathbb {T}}^n). \end{aligned}$$

(1)

Then we can check that

$$\begin{aligned} \int _{{\mathbb {T}}^m}(1_{{\mathbb {T}}^m},f)^*\Phi&=\sum _{i,j,k,l}h_{jk} P_i^jP_l^k\int _{{\mathbb {T}}^m} *_g\textrm{d}x^i\wedge \textrm{d}x^l\\&=\sum _{i,j,k,l}h_{jk} P_i^jP_l^k g^{il}\textrm{vol}_g({\mathbb {T}}^m)\\&=\textrm{tr}({}^tPG^{-1}PH)\textrm{vol}_g({\mathbb {T}}^m)=:\Vert P\Vert ^2\textrm{vol}_g({\mathbb {T}}^m)\ge 0. \end{aligned}$$

Consequently, by the positivity of $G^{-1}$ and H, $\int _{{\mathbb {T}}^m}(1_{{\mathbb {T}}^m},f)^*\Phi =0$ iff $P=0$.

Proposition 3.6

Assume that $f^*:H^1({\mathbb {T}}^n,{\mathbb {R}})\rightarrow H^1({\mathbb {T}}^m,{\mathbb {R}})$ is not the zero map. Then

• (i) $\Vert P\Vert ^{-1}\Phi$ is a $\sigma _1$-calibration,

• (ii) f is a $(\sigma _1,\Vert P\Vert ^{-1}\Phi )$-calibrated map if $f(x)=Px+a$ for some $a\in {\mathbb {T}}^m$,

• (iii) f minimizes $\mathcal {E}_2$ in its homotopy class iff $f(x)=Px+a$ for some $a\in {\mathbb {T}}^m$.

Proof

We fix $x\in {\mathbb {T}}^m$ and put $\textrm{d}f_x:=A=(A_i^j)\in M_{n,m}({\mathbb {R}})$, and show $(1_{{\mathbb {T}}^m},f)^*\Phi \le \sigma _1(f)\textrm{vol}_g$ at x. Since

$$\begin{aligned} (1_{{\mathbb {T}}^m},f)^*\Phi |_x&=\sum _{i,j,k,l}h_{jk} P_i^jA_l^k *_g\textrm{d}x^i\wedge \textrm{d}x^l|_x\\&=\left( \sum _{i,j,k,l}h_{jk} P_i^jA_l^k g^{il}\right) \textrm{vol}_g|_x\\&=\left( \textrm{tr}({}^tPG^{-1}AH)\right) \textrm{vol}_g|_x. \end{aligned}$$

Here, by the Cauchy–Schwarz inequality, we have

$$\begin{aligned} \textrm{tr}({}^tPG^{-1}AH)\le \sqrt{\Vert P\Vert \Vert A\Vert }, \end{aligned}$$

and the equality holds iff $A=\lambda P$ for a constant $\lambda \ge 0$. Therefore, we have

$$\begin{aligned} (1_{{\mathbb {T}}^m},f)^*\Phi&\le \Vert P\Vert \sigma _1(f)\textrm{vol}_g, \end{aligned}$$

which implies that $\Vert P\Vert ^{-1}\Phi$ is a $\sigma _1$-calibration. Moreover, the equality holds iff $\textrm{d}f_x=\lambda _x \cdot {}^t P$ for some $\lambda _x\ge 0$. Therefore, $f(x)={}^tPx+a$ for some $a\in {\mathbb {T}}^m$ is a $(\sigma _1,\Vert P\Vert ^{-1}\Phi )$-calibrated map.

For any $f\in C^\infty ({\mathbb {T}}^m,{\mathbb {T}}^n)$, we have

$$\begin{aligned} \int _X(1_{{\mathbb {T}}^m},f)^*\Phi&\le \Vert P\Vert \int _X\sigma _1(f)\textrm{vol}_g \le \Vert P\Vert \sqrt{\textrm{vol}_g({\mathbb {T}}^m)\mathcal {E}_2(f)} \end{aligned}$$

by the Cauchy–Schwartz inequality. Moreover, we have the following equality

$$\begin{aligned} \int _X(1_{{\mathbb {T}}^m},f)^*\Phi =\Vert P\Vert \sqrt{\textrm{vol}_g({\mathbb {T}}^m)\mathcal {E}_2(f)} \end{aligned}$$

iff $\textrm{d}f_x=\lambda _x\cdot {}^tP$ for some $\lambda _x\ge 0$ and $\sigma _1(f)$ is a constant function on ${\mathbb {T}}^m$. Since $\sigma _1(f)(x)=\lambda _x\Vert P\Vert$, if $\sigma _1(f)$ is constant, then $\lambda _x=\lambda$ is independent of x. Hence we may write $f(x)=\lambda \cdot {}^tPx+a$ for some $a\in {\mathbb {T}}^m$. Moreover, since $f^*=P$ on $H^1({\mathbb {T}}^n)$, we have $\lambda =1$. $\square$

In the above proposition, we cannot show that every $(\sigma _1,\Vert P\Vert ^{-1}\Phi )$-calibrated map is given by $f(x)={}^tPx+a$ for some a. We can give a counterexample as follows.

Suppose $m=n=1$ and let $P=1$. If we put $f(x)=x+\frac{1}{2\pi }\sin (2\pi x)$, then it gives a smooth map ${\mathbb {T}}^1\rightarrow {\mathbb {T}}^1$ homotopic to the identity map. Then one can check that f is a $(\sigma _1,\Vert P\Vert ^{-1}\Phi )$-calibrated map since $f'(x)\ge 0$.

4 The Lower Bound of p-Energy

In this section, we give the lower bound of p-energy in the general situation. Let (X, g) and (Y, h) be Riemannian manifolds and assume X is compact and oriented. Now we have the decomposition

$$\begin{aligned} \Lambda ^kT^*_{(x,y)}(X\times Y) \cong \bigoplus _{l=0}^k\Lambda ^lT^*_xX\otimes \Lambda ^{k-l}T^*_yY, \end{aligned}$$

then denote by $\Omega ^{l,k-l}(X\times Y)\subset \Omega ^k(X\times Y)$ the set consisting of smooth sections of $\Lambda ^lT^*_xX\otimes \Lambda ^{k-l}T^*_yY$. For $\Phi \in \Omega ^k(X\times Y)$, let $|\Phi _{(x,y)}|$ be the norm with respect to the metric $g\oplus h$ on $X\times Y$.

Lemma 4.1

Let $\Phi \in \Omega ^{m-k,k}(X\times Y)$ be closed and $\sup _{x,y}|\Phi _{(x,y)}|<\infty$. Then there is a constant $C>0$ depending only on $\Phi ,m,n,k$ such that $C\Phi$ is a $\sigma _k$-calibration.

Proof

Fix $x\in X$ and let $\{ e_1,\ldots ,e_m\}$ and $\{ e_1',\ldots ,e_n'\}$ be an orthonormal basis of $T_xX$ and $T_{f(x)}Y$, respectively. Put

$$\begin{aligned} \mathcal {I}_k^m :=\left\{ I=(i_1,\ldots ,i_k)\in {\mathbb {Z}}^k\left| \, 0\le i_1<\cdots <i_k\le m\right. \right\} . \end{aligned}$$

For $I=(i_1,\ldots ,i_k)\in \mathcal {I}_k^m$, $J=(j_1,\ldots ,j_k)\in \mathcal {I}_k^n$, we write

$$\begin{aligned} e_I:=e_{i_1}\wedge \cdots \wedge e_{i_k},\quad e_J':=e_{j_1}'\wedge \cdots \wedge e_{j_k}'. \end{aligned}$$

Then we have

$$\begin{aligned} \Phi _{(x,f(x))}=\sum _{I\in \mathcal {I}_k^m,J\in \mathcal {I}_k^n}\Phi _{IJ}(*_ge_I)\wedge e'_J \end{aligned}$$

for some $\Phi _{IJ}\in {\mathbb {R}}$ and

$$\begin{aligned} \left\{ (1_X,f)^*\Phi \right\} _x&=\sum _{I,J}\Phi _{IJ}(*_ge_I)\wedge \textrm{d}f_x^*e'_J. \end{aligned}$$

If we denote by $(\textrm{d}f_x)_{IJ}$ the $k\times k$ matrix whose (p, q)-component is given by $g(\textrm{d}f_x(e_{i_q}), e'_{j_p})$, then we have

$$\begin{aligned} (*_ge_I)\wedge \textrm{d}f_x^*e'_J&=\det ((\textrm{d}f_x)_{IJ})\textrm{vol}_g|_x\le k!|\textrm{d}f_x|^k\textrm{vol}_g|_x, \end{aligned}$$

therefore, we can see

$$\begin{aligned} \left\{ (1_X,f)^*\Phi \right\} _x&\le \left( \sum _{I,J}|\Phi _{IJ}|\right) k! |\textrm{d}f_x|^k\textrm{vol}_g|_x. \end{aligned}$$

Since $|\Phi _{x,f(x)}|^2=\sum _{I,J}|\Phi _{I,J}|^2$, we have

$$\begin{aligned} (1_X,f)^*\Phi&\le k! (\#\mathcal {I}_k^m)(\#\mathcal {I}_k^n)\sup _{x,y}|\Phi _{(x,y)}| \sigma _k(f)\textrm{vol}_g, \end{aligned}$$

which implies the assertion. $\square$

For $f\in C^\infty (X,Y)$, denote by $[f^*]^k$ the pullback $H^k(Y,{\mathbb {R}})\rightarrow H^k(X,{\mathbb {R}})$ of f. For a closed form $\alpha \in \Omega ^k(Y)$, denote by $[\alpha ]\in H^k(Y,{\mathbb {R}})$ its cohomology class. Put

$$\begin{aligned} H^k_{\textrm{bdd}}(Y,{\mathbb {R}}) :=\left\{ [\alpha ]\in H^k(Y,{\mathbb {R}})|\, \alpha \in \Omega ^k(Y),\, \textrm{d}\alpha =0,\, \sup _{y\in Y}h(\alpha _y,\alpha _y)<\infty \right\} . \end{aligned}$$

This is a subspace of $H^k(Y,{\mathbb {R}})$, and we have $H^k_{\textrm{bdd}}(Y,{\mathbb {R}})=H^k(Y,{\mathbb {R}})$ if Y is compact. Denote by $[f^*]^k_{\textrm{bdd}}$ the restriction of $[f^*]^k$ to $H^k_{\textrm{bdd}}(Y,{\mathbb {R}})$. Fixing a basis of $H^k(X,{\mathbb {R}})$ and $H^k_{\textrm{bdd}}(Y,{\mathbb {R}})$, we obtain the matrix $P=P([f^*]^k_{\textrm{bdd}})\in M_{N,d}({\mathbb {R}})$ of $[f^*]^k_{\textrm{bdd}}$, where $d=\dim H^k_{\textrm{bdd}}(Y,{\mathbb {R}})$ and $N=\dim H^k(X,{\mathbb {R}})$. Put $|P|:=\sqrt{\textrm{tr}({}^tPP)}$, which may depends on the choice of basis. Here, since d may become infinity, we may have $|P|=\infty$.

Theorem 4.2

Let $(X^m,g)$ and $(Y^n,h)$ be Riemannian manifolds and X be compact and oriented. For any $1\le k\le m$, there is a constant $C>0$ depending only on k, (X, g), and (Y, h) and the basis of $H^k(X,{\mathbb {R}})$, $H^k_{\textrm{bdd}}(Y,{\mathbb {R}})$ such that for any $f\in C^\infty (X,Y)$, we have

$$\begin{aligned} \mathcal {E}_k(f)\ge C|P([f^*]^k_{\textrm{bdd}})|. \end{aligned}$$

In particular, if $[f^*]^k_{\textrm{bdd}}$ is a nonzero map, then the infimum of $\mathcal {E}_k|_{[f]}$ is positive.

Proof

Take bounded closed k-forms $\beta _1,\ldots ,\beta _d\in \Omega ^k(Y)$ such that $\{ [\beta _l]\}_l$ is a basis of $H^k_{\textrm{bdd}}(Y,{\mathbb {R}})$.

By the Hodge Theory, $H^k(X)$ is isomorphic to the space of harmonic k-forms as vector spaces. Therefore, for any basis of $H^k(X,{\mathbb {R}})$, there is a corresponding basis $\alpha _1,\ldots ,\alpha _N\in \Omega ^k(X)$ of the space of harmonic k-forms. Let $G_{ij}:=\int _X\alpha _i\wedge *_g\alpha _j$, which is symmetric positive definite.

Define $P=(P_{ij})\in M_{N,d}({\mathbb {R}})$ by $[f^*]^k_{\textrm{bdd}}([\beta _j])=\sum _iP_{ij}[\alpha _i]$. If we put

$$\begin{aligned} \Phi :=\sum _{i,j} P_{ij}\beta _j\wedge (*_g\alpha _i), \end{aligned}$$

then every $\beta _j\wedge (*_g\alpha _i)$ is closed and satisfies the assumption of Lemma 4.1, since X is compact and $\beta _j$ is bounded. Take the constant $C_{ij}>0$ as in Lemma 4.1. Here, $C_{ij}$ is depending only on m, n, k, and $\alpha _i,\beta _j$. Then for any $f\in C^\infty (X,Y)$, we have

$$\begin{aligned} (1_X,f)^*\left\{ \beta _j\wedge (*_g\alpha _i)\right\}&\le C_{ij}\sigma _k(f)\textrm{vol}_g,\\ (1_X,f)^*\Phi&\le \sum _{i,j}C_{ij}|P_{ij}|\sigma _k(f)\textrm{vol}_g\\&\le \sqrt{\sum _{i,j}C_{ij}^2}|P|\sigma _k(f)\textrm{vol}_g, \end{aligned}$$

hence

$$\begin{aligned} \mathcal {E}_k(f)\ge \left( \sum _{i,j}C_{ij}^2\right) ^{-1/2} |P|^{-1}\int _X(1_X,f)^*\Phi . \end{aligned}$$

Moreover, we have

$$\begin{aligned} \int _X(1_X,f)^*\Phi&= \sum _{i,j} \int _X P_{ij}f^*\beta _j\wedge (*_g\alpha _i)\\&=\sum _{i,j} \int _X P_{ij}\sum _kP_{kj}\alpha _k\wedge (*_g\alpha _i)\\&=\sum _{i,j,k} P_{ij}P_{kj}G_{ki}. \end{aligned}$$

If we denote by $\lambda >0$ the minimum eigenvalue of $(G_{ij})_{i,j}$, then we have $\sum _{i,j,k} P_{ij}P_{kj}G_{ki}\ge \lambda |P|^2$. Hence we obtain

$$\begin{aligned} \mathcal {E}_k(f)\ge \lambda \left( \sum _{i,j}C_{ij}^2\right) ^{-1/2} |P|. \end{aligned}$$

$\square$

Remark 4.3

Combining the above theorem with Proposition 3.1, we also have the lower bound of $\mathcal {E}_p$ for any $p\ge k$.

5 Energy of the Identity Maps

In this section, we consider when the identity map on compact oriented Riemannian manifold X minimizes the energy. Here, we consider the family of energies. For Riemannian manifolds $(X^m,g)$, $(Y^n,h)$ and points $x\in X$, $y\in Y$, take a linear map $A:T_xX\rightarrow T_yY$. Fixing orthonormal basis of $T_xX$ and $T_yY$, we can regard A as an $n\times m$-matirx. Denote by $a_1,\ldots ,a_m\in {\mathbb {R}}_{\ge 0}$ the eigenvalues of ${}^tA\cdot A$, then put

$$\begin{aligned} |A|_p:=\left( \sum _{i=1}^m a_i^{p/2}\right) ^{1/p} \end{aligned}$$

for $p>0$. Then $|A|_p$ is independent of the choice of the orthonormal basis of $T_xX$. For a smooth map $f:X\rightarrow Y$, let

$$\begin{aligned} \sigma _{p,q}(f)|_x&:=|\textrm{d}f_x|_p^q,\\ \mathcal {E}_{p,q}(f)&:=\int _X\sigma _{p,q}(f)\textrm{vol}_g. \end{aligned}$$

Note that $\sigma _{2,p}=\sigma _p$ and $\mathcal {E}_{2,p}=\mathcal {E}_p$.

From now onward, we consider $(Y,h)=(X,g)$ and a map $f:X\rightarrow X$. Let $1_X$ be the identity map of X.

Proposition 5.1

If $1_X$ minimizes $\mathcal {E}_{p,q}|_{[1_X]}$, then it also minimizes $\mathcal {E}_{p',q'}|_{[1_X]}$ for any $p'\ge p$ and $q'\ge q$.

Proof

First of all, for any smooth map f, we have

$$\begin{aligned} |\textrm{d}f_x|_p&\le m^{1/p-1/p'}| \textrm{d}f_x|_{p'},\\ \mathcal {E}_{p,q}(f)&\le m^{q/p-q/p'}\textrm{vol}_g(X)^{1-q/q'} \left( \int _X|\textrm{d}f|_{p'}^{q'}\textrm{vol}_g\right) ^{q/q'}, \end{aligned}$$

by the Hölder’s inequality, which gives $\mathcal {E}_{p',q'}(f)\ge C\mathcal {E}_{p,q}(f)^{q'/q}$ for some constant $C>0$. Moreover, we have the equality for $f=1_X$. Therefore, we can see

$$\begin{aligned} \inf \mathcal {E}_{p',q'}|_{[1_X]} \ge \inf C\mathcal {E}_{p,q}^{q'/q}|_{[1_X]} = C\mathcal {E}_{p,q}(1_X)^{q'/q}= \mathcal {E}_{p',q'}(1_X) \ge \inf \mathcal {E}_{p',q'}|_{[1_X]}. \end{aligned}$$

$\square$

Proposition 5.2

(cf. [7, Lemma 2.2]) Let (X, g) be a compact oriented Riemannian manifold of dimension m. Then $1_X$ minimizes $\mathcal {E}_{1,m}$ in its homotopy class.

Proof

The proof is essentially given by [7, Lemma 2.2]. For any map $f:X\rightarrow X$, we can see

$$\begin{aligned} f^*\textrm{vol}_g=\det (\textrm{d}f)\textrm{vol}_g \le m^{-m}\sigma _{1,m}(f)\textrm{vol}_g. \end{aligned}$$

Here, the second inequality follows from the inequality

$$\begin{aligned} \frac{\sum _{i=1}^ma_i}{m}\ge \left( \prod _{i=1}^ma_i\right) ^{1/m} \end{aligned}$$

for any $a_i\ge 0$. Therefore, we can see

$$\begin{aligned} \mathcal {E}_{1,m}(f)\ge m^m\int _Xf^*\textrm{vol}_g. \end{aligned}$$

Moreover, the equality holds if $f=1_X$. $\square$

Next we consider the analogy of the above proposition. We assume that X has a nontrivial parallel k-form.

Denote by $g_0$ the standard metric on ${\mathbb {R}}^m$, which also induces the metric on $\Lambda ^k({\mathbb {R}}^m)^*$. Let $\varphi _0\in \Lambda ^k({\mathbb {R}}^m)^*$ and fix an orientation of ${\mathbb {R}}^m$. For a k-form $\varphi$ and a Riemannian metric g on an oriented manifold X, we say that $(g_0,\varphi _0)$ is a local model of $(g,\varphi )$ if for any $x\in X$ there is an orientation preserving isometry $I:{\mathbb {R}}^m\rightarrow T_xX$ such that $I^*(\varphi |_x)=\varphi _0$.

Denote by $*_{g_0}:\Lambda ^k({\mathbb {R}}^m)^*\rightarrow \Lambda ^{m-k}({\mathbb {R}}^m)^*$ the Hodge star operator induced by the standard metric and let $\textrm{vol}_{g_0}\in \Lambda ^m({\mathbb {R}}^m)^*$ be the volume form. First of all, we show the following proposition for the local model $(g_0,\varphi _0)$.

Proposition 5.3

Let $(g_0,\varphi _0)$ be as above. Assume that $\left| \iota _u\varphi _0\right| _{g_0}$ is independent of $u\in {\mathbb {R}}^m$ if $|u|_{g_0}=1$. We have

$$\begin{aligned} A^*\varphi _0\wedge *_{g_0} \varphi _0\le \frac{|\varphi _0|_{g_0}^2}{m}|A|_k^k\textrm{vol}_{g_0} \end{aligned}$$

for any $A\in M_m({\mathbb {R}})$. Moreover, if $A=\lambda T$ for $\lambda \in {\mathbb {R}}$, $T\in O(m)$ and $A^*\varphi _0=\lambda '\varphi _0$ for some $\lambda '\ge 0$, then we have the equality.

Proof

For any A, we can take oriented orthonormal basis $\{ e_1,\ldots ,e_m\}$ and $e'_1,\ldots ,e'_m$ of $({\mathbb {R}}^m)^*$ such that $A^*e'_i=a_ie_i$ for some $a_i\in {\mathbb {R}}$. We put

$$\begin{aligned} \varphi _0=\sum _{I\in \mathcal {I}_k^m}F_Ie_I=\sum _{I\in \mathcal {I}_k^m}F'_Ie'_I \end{aligned}$$

for some $F_I,F'_I\in {\mathbb {R}}$. Now, put $a_I:=a_{i_1}\cdots a_{i_k}$ for $I=(i_1,\ldots ,i_k)\in \mathcal {I}_k^m$. The we have $A^*\varphi _0=\sum _IF'_Ia_Ie_I$ and

$$\begin{aligned} A^*\varphi _0\wedge *_{g_0}\varphi _0 = g_0( A^*\varphi _0,\varphi _0)\textrm{vol}_{g_0}&=\sum _IF_IF'_Ia_I\textrm{vol}_{g_0}\\&\le \sum _I|F_IF'_I||a_I|\textrm{vol}_{g_0}. \end{aligned}$$

If we put $\{ I\}:=\{ i_1,\ldots ,i_k\}$, then

$$\begin{aligned} |a_I|=\left( |a_{i_1}|^k\cdots |a_{i_k}|^k\right) ^{1/k} \le \frac{1}{k} \sum _{j\in \{ I\}}|a_j|^k, \end{aligned}$$

therefore, we obtain

$$\begin{aligned} \sum _I|F_IF'_I||a_I|&\le \frac{1}{k}\sum _I |F_IF'_I|\sum _{j\in \{ I\}}|a_j|^k\\&= \frac{1}{k}\sum _{j=1}^m |a_j|^k\sum _{I\in \mathcal {I}_k^m,j\in \{ I\}} |F_IF'_I|. \end{aligned}$$

Denote by $\hat{g}_0:({\mathbb {R}}^m)^*\rightarrow {\mathbb {R}}^m$ the isomorphism induced by the metric $g_0$. Put

$$\begin{aligned} \varphi _1:=\sum _{I\in \mathcal {I}_k^m}|F_I|e_I,\quad \varphi _2:=\sum _{I\in \mathcal {I}_k^m}|F'_I|e_I \end{aligned}$$

and define an orthogonal matrix $U:{\mathbb {R}}^m\rightarrow {\mathbb {R}}^m$ by $U\circ \hat{g}_0(e_j)=\hat{g}_0(e'_j)$. Now we can see

$$\begin{aligned} \sum _{I\in \mathcal {I}_k^m,j\in \{ I\}} |F_IF'_I| =g_0\left( \iota _{\hat{g}_0(e_j)}\varphi _1,\iota _{\hat{g}_0(e_j)}\varphi _2\right)&\le \left| \iota _{\hat{g}_0(e_j)}\varphi _1\right| _{g_0} \cdot \left| \iota _{\hat{g}_0(e_j)}\varphi _2\right| _{g_0}\\&=\left| \iota _{\hat{g}_0(e_j)}\varphi _0\right| _{g_0} \cdot \left| \iota _{\hat{g}_0(e_j)}(U^*\varphi _0)\right| _{g_0} \end{aligned}$$

and

$$\begin{aligned} \left| \iota _{\hat{g}_0(e_j)}(U^*\varphi _0)\right| _{g_0} =\left| U^*(\iota _{U\circ \hat{g}_0(e_j)}\varphi _0)\right| _{g_0} =\left| \iota _{U\circ \hat{g}_0(e_j)}\varphi _0\right| _{g_0}. \end{aligned}$$

Then by the assumption, we can see that $C=\left| \iota _{\hat{g}_0(e_j)}\varphi _0\right| _{g_0}=\left| \iota _{U\circ \hat{g}_0(e_j)}\varphi _0\right| _{g_0}$ is independent of j, therefore, we have $\sum _{I\in \mathcal {I}_k^m,j\in \{ I\}} |F_IF'_I|\le C^2$ and

$$\begin{aligned} A^*\varphi _0\wedge *_{g_0}\varphi _0 \le \frac{C^2}{k}\sum _{j=1}^m |a_j|^k\textrm{vol}_{g_0}=\frac{C^2}{k}|A|_k^k\textrm{vol}_{g_0}. \end{aligned}$$

In the above inequalities, we have the equality if $A=1_m$, then we can determine the constant C. Moreover, we can also check that the equality holds if $A=\lambda T$, where $\lambda \in {\mathbb {R}}$, $T\in O(m)$ and $A^*\varphi _0=\lambda '\varphi _0$ for some $\lambda '\ge 0$. $\square$

Proposition 5.4

Let $(X^m,g)$ be a compact oriented Riemannian manifold and $\varphi \in \Omega ^k(X)$ be a harmonic form. Assume that there is a local model $(g_0,\varphi _0)$ of $(g,\varphi )$ and $|\iota _u\varphi _0|_{g_0}$ is independent of $u\in {\mathbb {R}}^m$ if $|u|_{g_0}=1$. Denote by $\textrm{pr}_i:X\times X\rightarrow X$ the projection to i-th component for $i=1,2$. Then $\Phi =m|\varphi _0|_{g_0}^{-2}\textrm{pr}_2^*\varphi \wedge \textrm{pr}_1^*(*_g\varphi )$ is an $\sigma _{k,k}$-calibration. Moreover, any isometry $f:X\rightarrow X$ with $f^*\varphi =\varphi$ is $(\sigma _{k,k},\Phi )$-calibrated.

Proof

$\Phi$ is an $\sigma _{k,k}$-calibration iff

$$\begin{aligned} f^*\varphi \wedge *_g\varphi \le \frac{|\varphi _0|_{g_0}^2}{m}|\textrm{d}f|_k^k\textrm{vol}_g. \end{aligned}$$

By putting $A=\textrm{d}f_x$ and identifying ${\mathbb {R}}^m\cong T_xX$, this is equivalent to the inequality in Proposition 5. Moreover, the equality holds if $(\textrm{d}f_x)^*\varphi |_x = \varphi |_x$ for all $x\in X$ and $\textrm{d}f_x$ is isometry. $\square$

Next we have to consider when the assumption for $(g_0,\varphi _0)$ is satisfied. If $G\subset SO(m)$ is a closed subgroup, then the linear action of SO(m) on ${\mathbb {R}}^m$ induces the action of G on ${\mathbb {R}}^m$. Similarly, since SO(m) acts on $\Lambda ^k({\mathbb {R}}^m)^*$ for all k, G also acts on them. Here, ${\mathbb {R}}^m$ is irreducible as a G-representation if any subspace $W\subset {\mathbb {R}}^m$ which is closed under the G-action is equal to ${\mathbb {R}}^m$ or $\{ 0\}$. For $\varphi _0\in \Lambda ^k({\mathbb {R}}^m)^*$, denote by $\textrm{Stab}(\varphi _0)\subset SO(m)$ the stabilizer of $\varphi _0$.

Lemma 5.5

Let G be a closed subgroup of SO(m) and assume that ${\mathbb {R}}^m$ is irreducible as a G-representation. Moreover, assume that $G\subset \textrm{Stab}(\varphi _0)$. Then $|\iota _u\varphi _0|_{g_0}$ is independent of $u\in {\mathbb {R}}^m$ if $|u|_{g_0}=1$.

Proof

Define a linear map $\Psi :{\mathbb {R}}^m\rightarrow \Lambda ^{k-1}({\mathbb {R}}^m)^*$ by $\Psi (u):=\iota _u\varphi _0$, then we can see $\Psi$ is G-equivariant map since the G-action preserves $\varphi _0$. Since the SO(m)-action on $\Lambda ^{k-1}({\mathbb {R}}^m)^*$ preserves the inner product, we can see

$$\begin{aligned} g_0( A\Psi (u),A\Psi (v) ) =g_0( \Psi (u),\Psi (v) ) \end{aligned}$$

for any $A\in G$ and $u,v\in {\mathbb {R}}^m$. Moreover, the left-hand side is equal to $g_0( \Psi ( Au),\Psi ( Av) )$ since $\Psi$ is G-equivariant.

Now, let $e_1,\ldots , e_m$ be the standard orthonormal basis of ${\mathbb {R}}^m$ and define the symmetric matrix $H=(H_{ij})_{i,j}$ by $H_{ij}:=g_0(\Psi (e_i),\Psi (e_j))$. Then by the above argument, we have ${}^tAHA=H$. Let $\lambda \in {\mathbb {R}}$ be any eigenvalue of H and denote by $V(\lambda )\subset {\mathbb {R}}^m$ the eigenspace associated with $\lambda$. Then we can see that $V(\lambda )$ is closed under the G-action, hence we have $V(\lambda )={\mathbb {R}}^m$ by the irreducibility, which implies

$$\begin{aligned} |\Psi ( Au)|_{g_0}^2=\lambda |u|_{g_0}^2 \end{aligned}$$

for all $u\in {\mathbb {R}}^m$ and $A\in G$. $\square$

Let $(X^m,g)$ be an oriented Riemannian manifold and denote by $\textrm{Hol}_g\subset SO(m)$ the holonomy group. We consider $(X,g,\varphi ,G,g_0,\varphi _0)$, where $\varphi \in \Omega ^k(X)$ is closed, $(g_0,\varphi _0)$ is a local model of $(g,\varphi )$, and G is a closed subgroup of SO(m) such that $\textrm{Hol}_g\subset G\subset \textrm{Stab}(\varphi _0)$. The followings are examples.

Table 1 Examples of $(X,g,\varphi ,G,g_0,\varphi _0)$

We can apply Proposition 5.4 and Lemma 5.5 to the above cases and obtain the following result.

Theorem 5.6

Let $(X,g,\varphi )$ be an oriented compact Riemannian manifold whose geometric structure is one of Table 1 and let $\Phi$ be as in Proposition 5.4. Then the identity map $1_X$ is a $(\sigma _{k,k},\Phi )$-calibrated map. In particular, $1_X$ minimizes $\mathcal {E}_{k,k}$ in its homotopy class.

6 Intersection of Smooth Maps

In [2], Croke and Fathi introduced the homotopy invariant of a smooth map $f:X\rightarrow Y$ which gives the lower bound to the 2-energy $\mathcal {E}_2$. In this section, we compare our invariant with the invariant in [2].

First of all, we review the intersection of smooth map introduced in [2]. Let (X, g) and (Y, h) be Riemannian manifolds and suppose X is compact. Here, we do not assume X is oriented, and we use the volume measure $\mu _g$ of g instead of the volume form.

Croke and Fathi defined the following quantity

$$\begin{aligned} i_f(g,h)=\lim _{t\rightarrow \infty }\frac{1}{t}\int _{S_g(X)}\phi _t(v)\textrm{d} \textrm{Liou}_g(v) \end{aligned}$$

for a smooth map $f:X\rightarrow Y$ and called it the intersection of f. Here, $\textrm{Liou}_g$ is the Liouville measure on the unit tangent bundle $S_g(X)$ and $\phi _t^f(v)=\phi _t(v)$ is the minimum length of all paths in Y homotopic with the fixed endpoints to

$$\begin{aligned} s\mapsto f(\gamma _v(s)),\quad 0\le s\le t, \end{aligned}$$

where $\gamma _v$ is the geodesic from $p\in X$ with $\gamma _v'(0)=v\in S_g(X)$.

Theorem 6.1

[2] For a smooth map $f:X\rightarrow Y$, the intersection $i_f(g,h)$ is homotopy invariant, that is, $i_f(g,h)=i_{f'}(g,h)$ if $[f']=[f]$. Moreover, for any f, we have

$$\begin{aligned} \int _X \sigma _2(f)\textrm{d}\mu _g \ge \frac{m}{V(S^{m-1})^2\mu _g(X)} i_f(g,h)^2, \end{aligned}$$

where $V(S^{m-1})$ is the volume of the unit sphere $S^{m-1}$ in ${\mathbb {R}}^m$.

First of all, we introduce the variant of $i_f(g,h)$ and improve the above theorem. We put

$$\begin{aligned} j_f(g,h):=\lim _{t\rightarrow \infty }\frac{1}{t^2}\int _{S_g(X)}\phi _t(v)^2\textrm{d} \textrm{Liou}_g(v). \end{aligned}$$

Theorem 6.2

For a smooth map $f:X\rightarrow Y$, $j_f(g,h)$ is homotopy invariant. Moreover, for any f, we have

$$\begin{aligned} \int _X \sigma _2(f)\textrm{d}\mu _g \ge \frac{m}{V(S^{m-1})} j_f(g,h), \end{aligned}$$

where the equality holds iff the image of the geodesic in X by f minimizes the length in its homotopy class with the fixed endpoints.

Proof

The proof is parallel to that of Theorem 6.1. The homotopy invariance of $j_f(g,h)$ is same as the case of $i_f(g,h)$. See the proof of [2, Lemma 1.3].

Next we show the inequality. Here we can see

$$\begin{aligned} \int _X\sigma _2(f)\textrm{d}\mu _g&=\frac{m}{V(S^{m-1})}\int _{S_g(X)} | \textrm{d}f(v) |_h^2\textrm{d}\textrm{Liou}_g(v). \end{aligned}$$

For $s\ge 0$, let $g_s:S_g(X)\rightarrow S_g(X)$ be the geodesic flow. Since $g_s$ preserves the Liouville measure, we can see

$$\begin{aligned} \int _{S_g(X)}|\textrm{d}f(v)|_h^2\textrm{d}\textrm{Liou}_g(v)&=\frac{1}{t}\int _{S_g(X)}\left( \int _0^t|\textrm{d}f(g_sv)|_h^2 ds\right) \textrm{d}\textrm{Liou}_g(v)\\&=\frac{1}{t}\int _{S_g(X)}\mathcal {E}_2(f\circ \gamma _v|_{[0,t]})\textrm{d}\textrm{Liou}_g(v), \end{aligned}$$

where $\mathcal {E}_2$ is the 2-energy of the curves in (Y, h). If L(c) is the length of c, then we have

$$\begin{aligned} \mathcal {E}_2(c) =\int _a^b|c'(s)|_h^2ds&\ge \frac{1}{b-a}\left( \int _a^b|c'(s)|_hds\right) ^2\\&= \frac{1}{b-a}\min _cL(c)^2, \end{aligned}$$

therefore

$$\begin{aligned} \int _{S_g(X)}|\textrm{d}f(v)|_h^2\textrm{d}\textrm{Liou}_g(v)&\ge \frac{1}{t^2}\int _{S_g(X)}\phi _t(v)^2\textrm{d}\textrm{Liou}_g(v) \end{aligned}$$

for any $t>0$. Consequently, we have the second assertion by considering $t\rightarrow \infty$. Finally, we consider the condition when

$$\begin{aligned} \int _{S_g(X)}|\textrm{d}f(v)|_h^2\textrm{d}\textrm{Liou}_g(v)&= \lim _{t\rightarrow \infty }\frac{1}{t^2}\int _{S_g(X)}\phi _t(v)^2\textrm{d}\textrm{Liou}_g(v) \end{aligned}$$

holds. To consider it, we show

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{1}{t^2}\int _{S_g(X)}\phi _t(v)^2\textrm{d}\textrm{Liou}_g(v) =\inf _{t>0}\frac{1}{t^2}\int _{S_g(X)}\phi _t(v)^2\textrm{d}\textrm{Liou}_g(v). \end{aligned}$$

(2)

By [2, Lemma 1.2], we have

$$\begin{aligned} \phi _{t+t'}(v)\le \phi _{t'}(g_tv)+\phi _t(v) \end{aligned}$$

for any $t,t'\ge 0$. Then by combining the Cauchy–Schwarz inequality, we have

$$\begin{aligned} \int _{S_g(X)}\phi _{t+t'}(v)^2\textrm{d}\textrm{Liou}_g(v)&\le \int _{S_g(X)}\phi _{t'}(g_tv)^2\textrm{d}\textrm{Liou}_g(v)\\&\quad +2\sqrt{ \int _{S_g(X)}\phi _{t'}(g_tv)^2\textrm{d}\textrm{Liou}_g(v) \int _{S_g(X)}\phi _t(v)^2\textrm{d}\textrm{Liou}_g(v) }\\&\quad +\int _{S_g(X)}\phi _t(v)^2\textrm{d}\textrm{Liou}_g(v). \end{aligned}$$

Since the Liouville measure is invariant under the geodesic flow, we can see$\int _{S_g(X)}\phi _{t'}(g_tv)^2\textrm{d}\textrm{Liou}_g(v) =\int _{S_g(X)}\phi _{t'}(v)^2\textrm{d}\textrm{Liou}_g(v)$, hence

$$\begin{aligned} \int _{S_g(X)}\phi _{t+t'}(v)^2\textrm{d}\textrm{Liou}_g(v)&\le \left( \sqrt{ \int _{S_g(X)}\phi _{t'}(v)^2\textrm{d}\textrm{Liou}_g(v)} + \sqrt{ \int _{S_g(X)}\phi _t(v)^2\textrm{d}\textrm{Liou}_g(v)} \right) ^2. \end{aligned}$$

If we put

$$\begin{aligned} P_t:=\sqrt{ \int _{S_g(X)}\phi _t(v)^2\textrm{d}\textrm{Liou}_g(v)}, \end{aligned}$$

then we have $P_{t+t'}\le P_t+P_{t'}$, hence

$$\begin{aligned} \inf _{t>0}\frac{P(t)}{t} =\lim _{t\rightarrow \infty }\frac{P(t)}{t}. \end{aligned}$$

Thus, we obtain (2).

Now, suppose

$$\begin{aligned} \int _X \sigma _2(f)\textrm{d}\mu _g = \frac{m}{V(S^{m-1})} j_f(g,h). \end{aligned}$$

By the above argument, we can see that $f\circ \gamma _v|_{[0,t]}$ is geodesic for any $v\in S_g(X)$ and $t>0$, and $L(f\circ \gamma _v|_{[0,t]})$ gives the minimum of

$$\begin{aligned} \left\{ L(c)|\, c \text{ is } \text{ homotopic } \text{ with } \text{ the } \text{ fixed } \text{ endpoints } \text{ to } f\circ \gamma _v|_{[0,t]}\right\} . \end{aligned}$$

$\square$

Remark 6.3

By the Cauchy–Schwarz inequality, we have

$$\begin{aligned} j_f(g,h)\ge \frac{i_f(g,h)^2}{\mu _g(X)V(S^{m-1})}, \end{aligned}$$

therefore, the inequality in Theorem 6.2 implies the inequality in Theorem 6.1.

Next we compute $j_f(g,h)$ in the case of flat tori and compare with the lower bound obtained by Proposition 3.6. Let $({\mathbb {T}}^m,g)$ and $({\mathbb {T}}^n,h)$ be as in Sect. 3.4 and take a coordinate x on ${\mathbb {T}}^m$ and y on ${\mathbb {T}}^n$ as in Sect. 3.4.

Proposition 6.4

Let $f:{\mathbb {T}}^m\rightarrow {\mathbb {T}}^n$ be a smooth map such that $f^*([\textrm{d}y^j])=\sum _iP_i^j[\textrm{d}x^i]$ for $P=(P_i^j)\in M_{m,n}({\mathbb {Z}})$. If we define $\Phi$ by (1) in Sect. 3.4, then we have

$$\begin{aligned} j_f(g,h)=\frac{V(S^{m-1})}{m}\int _{{\mathbb {T}}^m}(1_{{\mathbb {T}}^m},f)^*\Phi . \end{aligned}$$

Proof

First of all, we can see that f is homotopic to the map given by $x\mapsto Px$ for $x\in {\mathbb {T}}^m$, hence it suffices to show the equality by putting $f(x)={}^tPx$.

Since the image of the geodesic by f minimizes the length in its homotopy class with the fixed endpoints, then by Theorem 6.2, we have $\mathcal {E}_2(f)=\frac{m}{V(S^{m-1})}j_f(g,h)$. we can compute $\mathcal {E}_2(f)$ directly as

$$\begin{aligned} \mathcal {E}_2(f)=\int _{{\mathbb {T}}^m}|\textrm{d}f|^2\textrm{vol}_g =\sum _{i,j,k,l}h_{ij}P_k^iP_l^jg^{kl}\textrm{vol}_g({\mathbb {T}}^m) =\Vert P\Vert ^2\textrm{vol}_g({\mathbb {T}}^m). \end{aligned}$$

Moreover, by the computation in Sect. 3.4, we have shown that

$$\begin{aligned} \int _{{\mathbb {T}}^m}(1_{{\mathbb {T}}^m},f)^*\Phi =\Vert P\Vert ^2\textrm{vol}_g({\mathbb {T}}^m). \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{{\mathbb {T}}^m}(1_{{\mathbb {T}}^m},f)^*\Phi =\Vert P\Vert ^2\textrm{vol}_g({\mathbb {T}}^m) =\frac{m}{V(S^{m-1})}j_f(g,h). \end{aligned}$$

$\square$