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Asaf Shachar
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Consider the following functional:

$$ E_k(f)=\frac{1}{2}\int_{M} \| \bigwedge^k df\|^2 \text{Vol}_{M}.$$

Theorem:

The Euler-Lagrange equation of $E_2$, is $A(\phi)=0$, where $A(\phi) \in \Gamma(\phi^*\TN)$ is defined by

$$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)\big)\bigg).$$

$$h_{\phi^*TN}:\phi^*TN \otimes \Lambda_2(\phi^*TN) \to \phi^*TN$$ is a linear map, which depend on the metric on $\N$, and is defined precisely below (see eq $(6)$, and replace $W$ with $\phi^*TN$).

Proof:

Let $\phi$ be a map $\M \to \N$, and $\phi_t:\M \to \N$ as smooth family, where $\phi_0=\phi$ and $\frac{\partial \phi_t}{\partial t}|_{t=0}:=V \in \Gamma(\phi^*(\TN))$. Then

$$ \frac{d}{dt}|_{t=0}E(\phi_t)=\frac{1}{2}\int_{\M}\frac{\partial{}}{dt}|_{t=0} \| d\phi_t \wedge d\phi_t \|^2 \text{Vol}_{\M}= \int_{\M} \langle d\phi \wedge d\phi, \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\rangle \text{Vol}_{\M}. \tag{1}$$

It is well-known that $\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0}=\nabla^{\phi^*(TN)}V \in \Gamma(T^*\M \otimes \phi^*(\TN))$.

Now,

$$\bigg(\nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\bigg)(X,Y)=$$ $$(\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(X) \wedge d\phi(Y)+d\phi(X) \wedge (\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(Y)=$$

$$ \nabla V(X)\wedge d\phi(Y)+d\phi(X) \wedge \nabla V(Y)= $$

$$\big(\nabla V \wedge d\phi+d\phi \wedge \nabla V\big) (X,Y), \tag{2}$$

where $\nabla V \wedge d\phi+d\phi \wedge \nabla V \in \Omega^2\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$ is defined by the last equality.

Thus, we have obtained

$$ \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}= \nabla V \wedge d\phi+d\phi \wedge \nabla V. \tag{3}$$

Define also $\xi=V \wedge d\phi \in \Omega^1\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$.

Lemma: $d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)=\nabla V \wedge d\phi+d\phi \wedge \nabla V$.

Assuming the lemma, we combine equations $(1),(3)$ and get

$$ \frac{d}{dt}|_{t=0}E(\phi_t)=\int_{\M} \langle d\phi \wedge d\phi, d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)\rangle \text{Vol}_{\M}=\int_{\M} \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle \text{Vol}_{\M}. \tag{5}$$

To find the exact $E-L$ equations, one further step needs to be taken: $$V \to \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle=\langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),V \wedge d\phi\rangle$$ is a linear functional in $V$, so it can be expressed as $V \to \langle V, A(\phi) \rangle_{\phi^*\TN}$,

where $A(\phi) \in \Gamma(\phi^*\TN)$. The E-L equation is $A(\phi)=0$.

We now turn to finding an explicit expression for this "representation":

The corresponding pointwise linear algebra situation is this:

We have two oriented $d$-dimensional inner product spaces $V,W$, together with maps $A \in \Hom(V,W),B \in \Hom(V,\Lambda_2(W))$, and we look for a bilinearbilinear map $$\psi: \Hom(V,W) \otimes \Hom(V,\Lambda_2(W)) \to W,$$$$\psi: \Hom(V,W) \times \Hom(V,\Lambda_2(W)) \to W,$$ satisfying

$$ \langle w \wedge A,B \rangle_{\Hom(V,\Lambda_2(W))}=\langle w, \psi(A,B) \rangle_W \, \text{ for every $w \in W$}$$

Proposition: With the notation as above, $\psi(A,B)=h_W\big(\tr_{V} (A \otimes B)\big)$ where $h_W:W \otimes \Lambda_2(W) \to W$ is defined by the linear extension of

$$ \tilde w \otimes (w_1 \wedge w_2) \to \langle \tilde w,w_2 \rangle w_1-\langle \tilde w,w_1 \rangle w_2. \tag{6}$$

Note $A \otimes B \in V^* \otimes V^* \otimes W \otimes \Lambda_2(W)$, so $\tr_{V} (A \otimes B) \in W \otimes \Lambda_2(W)$.

Proof:

It suffices to prove this for $A,B$ "pure" tensors, i.e $A=\alpha \otimes \tilde w,B=\beta \otimes (w_1 \wedge w_2)$, where $\alpha,\beta \in V^*,\tilde w,w_1,w_2 \in W$.

Now, on the one hand

$$ \langle w \wedge A,B \rangle_{\Hom(V,\Lambda_2(W))}= \langle \alpha \otimes (w \wedge \tilde w) ,\beta \otimes (w_1 \wedge w_2) \rangle_{\Hom(V,\Lambda_2(W))}=$$

$$ \langle \alpha , \beta \rangle_{V^*} \langle w \wedge \tilde w ,w_1 \wedge w_2 \rangle_{\Lambda_2(W)}. $$

On the other hand

$$ \tr_{V} (A \otimes B)= \langle \alpha , \beta \rangle_{V^*} \tilde w \otimes (w_1 \wedge w_2).$$

Thus, it's enough to show

$$ \langle w \wedge \tilde w ,w_1 \wedge w_2 \rangle_{\Lambda_2(W)}=\langle w , h_W\big(\tilde w \otimes (w_1 \wedge w_2)\big) \rangle_W,$$

but this nows follows directly form the definition of the induced inner product on $\Lambda_2(W)$, and the definition of $h_W$ (see $(6)$).

Using the above proposition, we deduce that $$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)\big)\bigg).$$


Consider the following functional:

$$ E_k(f)=\frac{1}{2}\int_{M} \| \bigwedge^k df\|^2 \text{Vol}_{M}.$$

Theorem:

The Euler-Lagrange equation of $E_2$, is $A(\phi)=0$, where $A(\phi) \in \Gamma(\phi^*\TN)$ is defined by

$$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)\big)\bigg).$$

$$h_{\phi^*TN}:\phi^*TN \otimes \Lambda_2(\phi^*TN) \to \phi^*TN$$ is a linear map, which depend on the metric on $\N$, and is defined precisely below (see eq $(6)$, and replace $W$ with $\phi^*TN$).

Proof:

Let $\phi$ be a map $\M \to \N$, and $\phi_t:\M \to \N$ as smooth family, where $\phi_0=\phi$ and $\frac{\partial \phi_t}{\partial t}|_{t=0}:=V \in \Gamma(\phi^*(\TN))$. Then

$$ \frac{d}{dt}|_{t=0}E(\phi_t)=\frac{1}{2}\int_{\M}\frac{\partial{}}{dt}|_{t=0} \| d\phi_t \wedge d\phi_t \|^2 \text{Vol}_{\M}= \int_{\M} \langle d\phi \wedge d\phi, \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\rangle \text{Vol}_{\M}. \tag{1}$$

It is well-known that $\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0}=\nabla^{\phi^*(TN)}V \in \Gamma(T^*\M \otimes \phi^*(\TN))$.

Now,

$$\bigg(\nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\bigg)(X,Y)=$$ $$(\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(X) \wedge d\phi(Y)+d\phi(X) \wedge (\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(Y)=$$

$$ \nabla V(X)\wedge d\phi(Y)+d\phi(X) \wedge \nabla V(Y)= $$

$$\big(\nabla V \wedge d\phi+d\phi \wedge \nabla V\big) (X,Y), \tag{2}$$

where $\nabla V \wedge d\phi+d\phi \wedge \nabla V \in \Omega^2\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$ is defined by the last equality.

Thus, we have obtained

$$ \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}= \nabla V \wedge d\phi+d\phi \wedge \nabla V. \tag{3}$$

Define also $\xi=V \wedge d\phi \in \Omega^1\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$.

Lemma: $d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)=\nabla V \wedge d\phi+d\phi \wedge \nabla V$.

Assuming the lemma, we combine equations $(1),(3)$ and get

$$ \frac{d}{dt}|_{t=0}E(\phi_t)=\int_{\M} \langle d\phi \wedge d\phi, d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)\rangle \text{Vol}_{\M}=\int_{\M} \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle \text{Vol}_{\M}. \tag{5}$$

To find the exact $E-L$ equations, one further step needs to be taken: $$V \to \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle=\langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),V \wedge d\phi\rangle$$ is a linear functional in $V$, so it can be expressed as $V \to \langle V, A(\phi) \rangle_{\phi^*\TN}$,

where $A(\phi) \in \Gamma(\phi^*\TN)$. The E-L equation is $A(\phi)=0$.

We now turn to finding an explicit expression for this "representation":

The corresponding pointwise linear algebra situation is this:

We have two oriented $d$-dimensional inner product spaces $V,W$, together with maps $A \in \Hom(V,W),B \in \Hom(V,\Lambda_2(W))$, and we look for a bilinear map $$\psi: \Hom(V,W) \otimes \Hom(V,\Lambda_2(W)) \to W,$$ satisfying

$$ \langle w \wedge A,B \rangle_{\Hom(V,\Lambda_2(W))}=\langle w, \psi(A,B) \rangle_W \, \text{ for every $w \in W$}$$

Proposition: With the notation as above, $\psi(A,B)=h_W\big(\tr_{V} (A \otimes B)\big)$ where $h_W:W \otimes \Lambda_2(W) \to W$ is defined by the linear extension of

$$ \tilde w \otimes (w_1 \wedge w_2) \to \langle \tilde w,w_2 \rangle w_1-\langle \tilde w,w_1 \rangle w_2. \tag{6}$$

Note $A \otimes B \in V^* \otimes V^* \otimes W \otimes \Lambda_2(W)$, so $\tr_{V} (A \otimes B) \in W \otimes \Lambda_2(W)$.

Proof:

It suffices to prove this for $A,B$ "pure" tensors, i.e $A=\alpha \otimes \tilde w,B=\beta \otimes (w_1 \wedge w_2)$, where $\alpha,\beta \in V^*,\tilde w,w_1,w_2 \in W$.

Now, on the one hand

$$ \langle w \wedge A,B \rangle_{\Hom(V,\Lambda_2(W))}= \langle \alpha \otimes (w \wedge \tilde w) ,\beta \otimes (w_1 \wedge w_2) \rangle_{\Hom(V,\Lambda_2(W))}=$$

$$ \langle \alpha , \beta \rangle_{V^*} \langle w \wedge \tilde w ,w_1 \wedge w_2 \rangle_{\Lambda_2(W)}. $$

On the other hand

$$ \tr_{V} (A \otimes B)= \langle \alpha , \beta \rangle_{V^*} \tilde w \otimes (w_1 \wedge w_2).$$

Thus, it's enough to show

$$ \langle w \wedge \tilde w ,w_1 \wedge w_2 \rangle_{\Lambda_2(W)}=\langle w , h_W\big(\tilde w \otimes (w_1 \wedge w_2)\big) \rangle_W,$$

but this nows follows directly form the definition of the induced inner product on $\Lambda_2(W)$, and the definition of $h_W$ (see $(6)$).

Using the above proposition, we deduce that $$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)\big)\bigg).$$


Consider the following functional:

$$ E_k(f)=\frac{1}{2}\int_{M} \| \bigwedge^k df\|^2 \text{Vol}_{M}.$$

Theorem:

The Euler-Lagrange equation of $E_2$, is $A(\phi)=0$, where $A(\phi) \in \Gamma(\phi^*\TN)$ is defined by

$$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)\big)\bigg).$$

$$h_{\phi^*TN}:\phi^*TN \otimes \Lambda_2(\phi^*TN) \to \phi^*TN$$ is a linear map, which depend on the metric on $\N$, and is defined precisely below (see eq $(6)$, and replace $W$ with $\phi^*TN$).

Proof:

Let $\phi$ be a map $\M \to \N$, and $\phi_t:\M \to \N$ as smooth family, where $\phi_0=\phi$ and $\frac{\partial \phi_t}{\partial t}|_{t=0}:=V \in \Gamma(\phi^*(\TN))$. Then

$$ \frac{d}{dt}|_{t=0}E(\phi_t)=\frac{1}{2}\int_{\M}\frac{\partial{}}{dt}|_{t=0} \| d\phi_t \wedge d\phi_t \|^2 \text{Vol}_{\M}= \int_{\M} \langle d\phi \wedge d\phi, \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\rangle \text{Vol}_{\M}. \tag{1}$$

It is well-known that $\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0}=\nabla^{\phi^*(TN)}V \in \Gamma(T^*\M \otimes \phi^*(\TN))$.

Now,

$$\bigg(\nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\bigg)(X,Y)=$$ $$(\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(X) \wedge d\phi(Y)+d\phi(X) \wedge (\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(Y)=$$

$$ \nabla V(X)\wedge d\phi(Y)+d\phi(X) \wedge \nabla V(Y)= $$

$$\big(\nabla V \wedge d\phi+d\phi \wedge \nabla V\big) (X,Y), \tag{2}$$

where $\nabla V \wedge d\phi+d\phi \wedge \nabla V \in \Omega^2\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$ is defined by the last equality.

Thus, we have obtained

$$ \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}= \nabla V \wedge d\phi+d\phi \wedge \nabla V. \tag{3}$$

Define also $\xi=V \wedge d\phi \in \Omega^1\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$.

Lemma: $d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)=\nabla V \wedge d\phi+d\phi \wedge \nabla V$.

Assuming the lemma, we combine equations $(1),(3)$ and get

$$ \frac{d}{dt}|_{t=0}E(\phi_t)=\int_{\M} \langle d\phi \wedge d\phi, d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)\rangle \text{Vol}_{\M}=\int_{\M} \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle \text{Vol}_{\M}. \tag{5}$$

To find the exact $E-L$ equations, one further step needs to be taken: $$V \to \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle=\langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),V \wedge d\phi\rangle$$ is a linear functional in $V$, so it can be expressed as $V \to \langle V, A(\phi) \rangle_{\phi^*\TN}$,

where $A(\phi) \in \Gamma(\phi^*\TN)$. The E-L equation is $A(\phi)=0$.

We now turn to finding an explicit expression for this "representation":

The corresponding pointwise linear algebra situation is this:

We have two oriented $d$-dimensional inner product spaces $V,W$, together with maps $A \in \Hom(V,W),B \in \Hom(V,\Lambda_2(W))$, and we look for a bilinear map $$\psi: \Hom(V,W) \times \Hom(V,\Lambda_2(W)) \to W,$$ satisfying

$$ \langle w \wedge A,B \rangle_{\Hom(V,\Lambda_2(W))}=\langle w, \psi(A,B) \rangle_W \, \text{ for every $w \in W$}$$

Proposition: With the notation as above, $\psi(A,B)=h_W\big(\tr_{V} (A \otimes B)\big)$ where $h_W:W \otimes \Lambda_2(W) \to W$ is defined by the linear extension of

$$ \tilde w \otimes (w_1 \wedge w_2) \to \langle \tilde w,w_2 \rangle w_1-\langle \tilde w,w_1 \rangle w_2. \tag{6}$$

Note $A \otimes B \in V^* \otimes V^* \otimes W \otimes \Lambda_2(W)$, so $\tr_{V} (A \otimes B) \in W \otimes \Lambda_2(W)$.

Proof:

It suffices to prove this for $A,B$ "pure" tensors, i.e $A=\alpha \otimes \tilde w,B=\beta \otimes (w_1 \wedge w_2)$, where $\alpha,\beta \in V^*,\tilde w,w_1,w_2 \in W$.

Now, on the one hand

$$ \langle w \wedge A,B \rangle_{\Hom(V,\Lambda_2(W))}= \langle \alpha \otimes (w \wedge \tilde w) ,\beta \otimes (w_1 \wedge w_2) \rangle_{\Hom(V,\Lambda_2(W))}=$$

$$ \langle \alpha , \beta \rangle_{V^*} \langle w \wedge \tilde w ,w_1 \wedge w_2 \rangle_{\Lambda_2(W)}. $$

On the other hand

$$ \tr_{V} (A \otimes B)= \langle \alpha , \beta \rangle_{V^*} \tilde w \otimes (w_1 \wedge w_2).$$

Thus, it's enough to show

$$ \langle w \wedge \tilde w ,w_1 \wedge w_2 \rangle_{\Lambda_2(W)}=\langle w , h_W\big(\tilde w \otimes (w_1 \wedge w_2)\big) \rangle_W,$$

but this nows follows directly form the definition of the induced inner product on $\Lambda_2(W)$, and the definition of $h_W$ (see $(6)$).

Using the above proposition, we deduce that $$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)\big)\bigg).$$


Omitted unnecessary parts. Made the presentation clearer and more succinct.
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Asaf Shachar
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Summary: I guess $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ is the Euler-Lagrange equation of the functional

$$ E(f)=\frac{1}{2}\int_{M} \| \bigwedge^k df\|^2 \text{Vol}_{M}.$$

Is it true? If not, what is Consider the E-L equation of thisfollowing functional?:

Edit: It turns out this is not the exact $E-L$ equation. However, it is true that if $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ then $f$ is a critical point of $E$. (See details below).$$ E_k(f)=\frac{1}{2}\int_{M} \| \bigwedge^k df\|^2 \text{Vol}_{M}.$$


 

In the case $k=2$, theThe Euler-Lagrange equationsequation of $E$$E_2$, areis $A(\phi)=0$, where $A(\phi) \in \Gamma(\phi^*\TN)$ is defined by

$$h_{\phi^*TN}:\phi^*TN \otimes \Lambda_2(\phi^*TN) \to \phi^*TN$$ is a linear map, which depend on the metric on $\N$, and is defined precisely below (see eq $(6)$, and replace $W$ with $\phi^*TN$).

Now, on can easily prove that

where $\nabla V \wedge d\phi+d\phi \wedge \nabla V \in \Omega^2\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$ is defined by the last equality, i.e it's the map $$(X,Y) \to \nabla V(X)\wedge d\phi(Y)+d\phi(X) \wedge \nabla V(Y).$$ Note that each of the two summands is not a $2$-form, but only the "sum" is. (Each summand is not anti-symmetric).

Thus, if $ \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)=0 $, then $\phi$ is a critical point of $E$.

The other direction might be false, (this can be seen by counting degrees of freedom, i.e $V \in \Gamma(\phi^*\TN)$ has $d$ degrees of freedom, whlie the equation $\delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)=0 $ is in fact a system of $d \cdot {d \choose 2}$ scalar equations).

To find the exact $E-L$ equations, one further step needs to be taken:

Consider $V \to \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle=\langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),V \wedge d\phi\rangle$ as $$V \to \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle=\langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),V \wedge d\phi\rangle$$ is a linear functional in $V$. As such, so it can be expressed (Riesz theorem) as $V \to \langle V, A(\phi) \rangle_{\phi^*\TN}$,

where $A(\phi) \in \Gamma(\phi^*\TN)$ is a section associated with $\phi$, which vanishes when $\delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)=0$. The EulerE-Lagrange equationsL equation is $A(\phi)=0$.

Can we find "explicitly" this section $A(\phi)$?We now turn to finding an explicit expression for this "representation":


 

Here is an attempt to do so: The corresponding pointwise linear algebra situation is this:

Using the above proposition, we deduce that $$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)\big)\bigg).$$

Can we find a more nice expression for $h_{\phi^*TN}$ ($h_W$)?

Summary: I guess $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ is the Euler-Lagrange equation of the functional

$$ E(f)=\frac{1}{2}\int_{M} \| \bigwedge^k df\|^2 \text{Vol}_{M}.$$

Is it true? If not, what is the E-L equation of this functional?

Edit: It turns out this is not the exact $E-L$ equation. However, it is true that if $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ then $f$ is a critical point of $E$. (See details below).


 

In the case $k=2$, the Euler-Lagrange equations of $E$, are $A(\phi)=0$, where $A(\phi) \in \Gamma(\phi^*\TN)$ is defined by

$$h_{\phi^*TN}:\phi^*TN \otimes \Lambda_2(\phi^*TN) \to \phi^*TN$$ is a linear map, which depend on the metric on $\N$, and is defined precisely below (see eq $(6)$).

Now, on can easily prove that

where $\nabla V \wedge d\phi+d\phi \wedge \nabla V \in \Omega^2\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$ is defined by the last equality, i.e it's the map $$(X,Y) \to \nabla V(X)\wedge d\phi(Y)+d\phi(X) \wedge \nabla V(Y).$$ Note that each of the two summands is not a $2$-form, but only the "sum" is. (Each summand is not anti-symmetric).

Thus, if $ \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)=0 $, then $\phi$ is a critical point of $E$.

The other direction might be false, (this can be seen by counting degrees of freedom, i.e $V \in \Gamma(\phi^*\TN)$ has $d$ degrees of freedom, whlie the equation $\delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)=0 $ is in fact a system of $d \cdot {d \choose 2}$ scalar equations).

To find the exact $E-L$ equations, one further step needs to be taken:

Consider $V \to \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle=\langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),V \wedge d\phi\rangle$ as a linear functional in $V$. As such, it can be expressed (Riesz theorem) as $V \to \langle V, A(\phi) \rangle_{\phi^*\TN}$,

where $A(\phi) \in \Gamma(\phi^*\TN)$ is a section associated with $\phi$, which vanishes when $\delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)=0$. The Euler-Lagrange equations is $A(\phi)=0$.

Can we find "explicitly" this section $A(\phi)$?


 

Here is an attempt to do so: The corresponding pointwise linear algebra situation is this:

Using the above proposition, we deduce that $$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)\big)\bigg).$$

Can we find a more nice expression for $h_{\phi^*TN}$ ($h_W$)?

Consider the following functional:

$$ E_k(f)=\frac{1}{2}\int_{M} \| \bigwedge^k df\|^2 \text{Vol}_{M}.$$

The Euler-Lagrange equation of $E_2$, is $A(\phi)=0$, where $A(\phi) \in \Gamma(\phi^*\TN)$ is defined by

$$h_{\phi^*TN}:\phi^*TN \otimes \Lambda_2(\phi^*TN) \to \phi^*TN$$ is a linear map, which depend on the metric on $\N$, and is defined precisely below (see eq $(6)$, and replace $W$ with $\phi^*TN$).

Now,

where $\nabla V \wedge d\phi+d\phi \wedge \nabla V \in \Omega^2\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$ is defined by the last equality.

To find the exact $E-L$ equations, one further step needs to be taken: $$V \to \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle=\langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),V \wedge d\phi\rangle$$ is a linear functional in $V$, so it can be expressed as $V \to \langle V, A(\phi) \rangle_{\phi^*\TN}$,

where $A(\phi) \in \Gamma(\phi^*\TN)$. The E-L equation is $A(\phi)=0$.

We now turn to finding an explicit expression for this "representation":

The corresponding pointwise linear algebra situation is this:

Using the above proposition, we deduce that $$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)\big)\bigg).$$

Added an attempt to make the E-L equation more explicit
Source Link
Asaf Shachar
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Summary: I guess $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ is the Euler-Lagrange equation of the functional

$$ E(f)=\frac{1}{2}\int_{M} \| \bigwedge^k df\|^2 \text{Vol}_{M}.$$

Is it true? If not, what is the E-L equation of this functional?

Edit: It turns out this is not the exact $E-L$ equation. However, it is true that if $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ then $f$ is a critical point of $E$. (See details below).


Theorem:

In the case $k=2$, the Euler-Lagrange equations of $E$, are $A(\phi)=0$, where $A(\phi) \in \Gamma(\phi^*\TN)$ is defined by

$$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)\big)\bigg).$$

$$h_{\phi^*TN}:\phi^*TN \otimes \Lambda_2(\phi^*TN) \to \phi^*TN$$ is a linear map, which depend on the metric on $\N$, and is defined precisely below (see eq $(6)$).

Here is an attempt to prove this for $k=2$:Proof:

Let $\phi$ be a map $\M \to \N$, and $\phi_t:\M \to \N$ as smooth family, where $\phi_0=\phi$ and $\frac{\partial \phi_t}{\partial t}|_{t=0}:=V \in \Gamma(\phi^*(\TN))$. Then

$$ \frac{d}{dt}|_{t=0}E(\phi_t)=\frac{1}{2}\int_{\M}\frac{\partial{}}{dt}|_{t=0} \| d\phi_t \wedge d\phi_t \|^2 \text{Vol}_{\M}= \int_{\M} \langle d\phi \wedge d\phi, \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\rangle \text{Vol}_{\M}. \tag{1}$$

It is well-known that $\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0}=\nabla^{\phi^*(TN)}V \in \Gamma(T^*\M \otimes \phi^*(\TN))$.

Now, on can easily prove that

$$\bigg(\nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\bigg)(X,Y)=$$ $$(\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(X) \wedge d\phi(Y)+d\phi(X) \wedge (\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(Y)=$$

$$ \nabla V(X)\wedge d\phi(Y)+d\phi(X) \wedge \nabla V(Y)= $$

$$\big(\nabla V \wedge d\phi+d\phi \wedge \nabla V\big) (X,Y), \tag{2}$$

where $\nabla V \wedge d\phi+d\phi \wedge \nabla V \in \Omega^2\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$ is defined by the last equality, i.e it's the map $$(X,Y) \to \nabla V(X)\wedge d\phi(Y)+d\phi(X) \wedge \nabla V(Y).$$ Note that each of the two summands is not a $2$-form, but only the "sum" is. (Each summand is not anti-symmetric).

Thus, we have obtained

$$ \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}= \nabla V \wedge d\phi+d\phi \wedge \nabla V. \tag{3}$$

Define also $\xi=V \wedge d\phi \in \Omega^1\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$.

Lemma: $d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)=\nabla V \wedge d\phi+d\phi \wedge \nabla V$.

Assuming the lemma, we combine equations $(1),(3)$ and get

$$ \frac{d}{dt}|_{t=0}E(\phi_t)=\int_{\M} \langle d\phi \wedge d\phi, d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)\rangle \text{Vol}_{\M}=\int_{\M} \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle \text{Vol}_{\M}. \tag{5}$$

Thus, if $ \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)=0 $, then $\phi$ is a critical point of $E$.

The other direction might be false, (this can be seen by counting degrees of freedom, i.e $V \in \Gamma(\phi^*\TN)$ has $d$ degrees of freedom, whlie the equation $\delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)=0 $ is in fact a system of $d \cdot {d \choose 2}$ scalar equations).

To find the exact $E-L$ equations, one further step needs to be taken:

Consider $V \to \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle=\langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),V \wedge d\phi\rangle$ as a linear functional in $V$. As such, it can be expressed (Riesz theorem) as $V \to \langle V, A(\phi) \rangle_{\phi^*\TN}$,

where $A(\phi) \in \Gamma(\phi^*\TN)$ is a section associated with $\phi$, which vanishes when $\delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)=0$. The Euler-Lagrange equations is $A(\phi)=0$.

Can we find "explicitly" this section $A(\phi)$?


Here is an attempt to do so: The corresponding pointwise linear algebra situation is this:

We have two oriented $d$-dimensional inner product spaces $V,W$, together with maps $A \in \Hom(V,W),B \in \Hom(V,\Lambda_2(W))$, and we look for a bilinear map $$\psi: \Hom(V,W) \otimes \Hom(V,\Lambda_2(W)) \to W,$$ satisfying

$$ \langle w \wedge A,B \rangle_{\Hom(V,\Lambda_2(W))}=\langle w, \psi(A,B) \rangle_W \, \text{ for every $w \in W$}$$

Proposition: With the notation as above, $\psi(A,B)=h_W\big(\tr_{V} (A \otimes B)\big)$ where $h_W:W \otimes \Lambda_2(W) \to W$ is defined by the linear extension of

$$ \tilde w \otimes (w_1 \wedge w_2) \to \langle \tilde w,w_2 \rangle w_1-\langle \tilde w,w_1 \rangle w_2. \tag{6}$$

Note $A \otimes B \in V^* \otimes V^* \otimes W \otimes \Lambda_2(W)$, so $\tr_{V} (A \otimes B) \in W \otimes \Lambda_2(W)$.

Proof:

It suffices to prove this for $A,B$ "pure" tensors, i.e $A=\alpha \otimes \tilde w,B=\beta \otimes (w_1 \wedge w_2)$, where $\alpha,\beta \in V^*,\tilde w,w_1,w_2 \in W$.

Now, on the one hand

$$ \langle w \wedge A,B \rangle_{\Hom(V,\Lambda_2(W))}= \langle \alpha \otimes (w \wedge \tilde w) ,\beta \otimes (w_1 \wedge w_2) \rangle_{\Hom(V,\Lambda_2(W))}=$$

$$ \langle \alpha , \beta \rangle_{V^*} \langle w \wedge \tilde w ,w_1 \wedge w_2 \rangle_{\Lambda_2(W)}. $$

On the other hand

$$ \tr_{V} (A \otimes B)= \langle \alpha , \beta \rangle_{V^*} \tilde w \otimes (w_1 \wedge w_2).$$

Thus, it's enough to show

$$ \langle w \wedge \tilde w ,w_1 \wedge w_2 \rangle_{\Lambda_2(W)}=\langle w , h_W\big(\tilde w \otimes (w_1 \wedge w_2)\big) \rangle_W,$$

but this nows follows directly form the definition of the induced inner product on $\Lambda_2(W)$, and the definition of $h_W$ (see $(6)$).

Using the above proposition, we deduce that $$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)\big)\bigg).$$

Can we find a more nice expression for $h_{\phi^*TN}$ ($h_W$)?


Summary: I guess $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ is the Euler-Lagrange equation of the functional

$$ E(f)=\frac{1}{2}\int_{M} \| \bigwedge^k df\|^2 \text{Vol}_{M}.$$

Is it true? If not, what is the E-L equation of this functional?

Edit: It turns out this is not the exact $E-L$ equation. However, it is true that if $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ then $f$ is a critical point of $E$. (See details below).


Here is an attempt to prove this for $k=2$:

Let $\phi$ be a map $\M \to \N$, and $\phi_t:\M \to \N$ as smooth family, where $\phi_0=\phi$ and $\frac{\partial \phi_t}{\partial t}|_{t=0}:=V \in \Gamma(\phi^*(\TN))$. Then

$$ \frac{d}{dt}|_{t=0}E(\phi_t)=\frac{1}{2}\int_{\M}\frac{\partial{}}{dt}|_{t=0} \| d\phi_t \wedge d\phi_t \|^2 \text{Vol}_{\M}= \int_{\M} \langle d\phi \wedge d\phi, \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\rangle \text{Vol}_{\M}. \tag{1}$$

It is well-known that $\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0}=\nabla^{\phi^*(TN)}V \in \Gamma(T^*\M \otimes \phi^*(\TN))$.

Now, on can easily prove that

$$\bigg(\nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\bigg)(X,Y)=$$ $$(\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(X) \wedge d\phi(Y)+d\phi(X) \wedge (\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(Y)=$$

$$ \nabla V(X)\wedge d\phi(Y)+d\phi(X) \wedge \nabla V(Y)= $$

$$\big(\nabla V \wedge d\phi+d\phi \wedge \nabla V\big) (X,Y), \tag{2}$$

where $\nabla V \wedge d\phi+d\phi \wedge \nabla V \in \Omega^2\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$ is defined by the last equality, i.e it's the map $$(X,Y) \to \nabla V(X)\wedge d\phi(Y)+d\phi(X) \wedge \nabla V(Y).$$ Note that each of the two summands is not a $2$-form, but only the "sum" is. (Each summand is not anti-symmetric).

Thus, we have obtained

$$ \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}= \nabla V \wedge d\phi+d\phi \wedge \nabla V. \tag{3}$$

Define also $\xi=V \wedge d\phi \in \Omega^1\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$.

Lemma: $d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)=\nabla V \wedge d\phi+d\phi \wedge \nabla V$.

Assuming the lemma, we combine equations $(1),(3)$ and get

$$ \frac{d}{dt}|_{t=0}E(\phi_t)=\int_{\M} \langle d\phi \wedge d\phi, d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)\rangle \text{Vol}_{\M}=\int_{\M} \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle \text{Vol}_{\M}. \tag{5}$$

Thus, if $ \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)=0 $, then $\phi$ is a critical point of $E$.

The other direction might be false, (this can be seen by counting degrees of freedom, i.e $V \in \Gamma(\phi^*\TN)$ has $d$ degrees of freedom, whlie the equation $\delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)=0 $ is in fact a system of $d \cdot {d \choose 2}$ scalar equations).

To find the exact $E-L$ equations, one further step needs to be taken:

Consider $V \to \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle=\langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),V \wedge d\phi\rangle$ as a linear functional in $V$. As such, it can be expressed (Riesz theorem) as $V \to \langle V, A(\phi) \rangle_{\phi^*\TN}$,

where $A(\phi) \in \Gamma(\phi^*\TN)$ is a section associated with $\phi$, which vanishes when $\delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)=0$.

Can we find "explicitly" this section $A(\phi)$?

Summary: I guess $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ is the Euler-Lagrange equation of the functional

$$ E(f)=\frac{1}{2}\int_{M} \| \bigwedge^k df\|^2 \text{Vol}_{M}.$$

Is it true? If not, what is the E-L equation of this functional?

Edit: It turns out this is not the exact $E-L$ equation. However, it is true that if $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ then $f$ is a critical point of $E$. (See details below).


Theorem:

In the case $k=2$, the Euler-Lagrange equations of $E$, are $A(\phi)=0$, where $A(\phi) \in \Gamma(\phi^*\TN)$ is defined by

$$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)\big)\bigg).$$

$$h_{\phi^*TN}:\phi^*TN \otimes \Lambda_2(\phi^*TN) \to \phi^*TN$$ is a linear map, which depend on the metric on $\N$, and is defined precisely below (see eq $(6)$).

Proof:

Let $\phi$ be a map $\M \to \N$, and $\phi_t:\M \to \N$ as smooth family, where $\phi_0=\phi$ and $\frac{\partial \phi_t}{\partial t}|_{t=0}:=V \in \Gamma(\phi^*(\TN))$. Then

$$ \frac{d}{dt}|_{t=0}E(\phi_t)=\frac{1}{2}\int_{\M}\frac{\partial{}}{dt}|_{t=0} \| d\phi_t \wedge d\phi_t \|^2 \text{Vol}_{\M}= \int_{\M} \langle d\phi \wedge d\phi, \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\rangle \text{Vol}_{\M}. \tag{1}$$

It is well-known that $\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0}=\nabla^{\phi^*(TN)}V \in \Gamma(T^*\M \otimes \phi^*(\TN))$.

Now, on can easily prove that

$$\bigg(\nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\bigg)(X,Y)=$$ $$(\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(X) \wedge d\phi(Y)+d\phi(X) \wedge (\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(Y)=$$

$$ \nabla V(X)\wedge d\phi(Y)+d\phi(X) \wedge \nabla V(Y)= $$

$$\big(\nabla V \wedge d\phi+d\phi \wedge \nabla V\big) (X,Y), \tag{2}$$

where $\nabla V \wedge d\phi+d\phi \wedge \nabla V \in \Omega^2\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$ is defined by the last equality, i.e it's the map $$(X,Y) \to \nabla V(X)\wedge d\phi(Y)+d\phi(X) \wedge \nabla V(Y).$$ Note that each of the two summands is not a $2$-form, but only the "sum" is. (Each summand is not anti-symmetric).

Thus, we have obtained

$$ \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}= \nabla V \wedge d\phi+d\phi \wedge \nabla V. \tag{3}$$

Define also $\xi=V \wedge d\phi \in \Omega^1\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$.

Lemma: $d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)=\nabla V \wedge d\phi+d\phi \wedge \nabla V$.

Assuming the lemma, we combine equations $(1),(3)$ and get

$$ \frac{d}{dt}|_{t=0}E(\phi_t)=\int_{\M} \langle d\phi \wedge d\phi, d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)\rangle \text{Vol}_{\M}=\int_{\M} \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle \text{Vol}_{\M}. \tag{5}$$

Thus, if $ \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)=0 $, then $\phi$ is a critical point of $E$.

The other direction might be false, (this can be seen by counting degrees of freedom, i.e $V \in \Gamma(\phi^*\TN)$ has $d$ degrees of freedom, whlie the equation $\delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)=0 $ is in fact a system of $d \cdot {d \choose 2}$ scalar equations).

To find the exact $E-L$ equations, one further step needs to be taken:

Consider $V \to \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle=\langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),V \wedge d\phi\rangle$ as a linear functional in $V$. As such, it can be expressed (Riesz theorem) as $V \to \langle V, A(\phi) \rangle_{\phi^*\TN}$,

where $A(\phi) \in \Gamma(\phi^*\TN)$ is a section associated with $\phi$, which vanishes when $\delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)=0$. The Euler-Lagrange equations is $A(\phi)=0$.

Can we find "explicitly" this section $A(\phi)$?


Here is an attempt to do so: The corresponding pointwise linear algebra situation is this:

We have two oriented $d$-dimensional inner product spaces $V,W$, together with maps $A \in \Hom(V,W),B \in \Hom(V,\Lambda_2(W))$, and we look for a bilinear map $$\psi: \Hom(V,W) \otimes \Hom(V,\Lambda_2(W)) \to W,$$ satisfying

$$ \langle w \wedge A,B \rangle_{\Hom(V,\Lambda_2(W))}=\langle w, \psi(A,B) \rangle_W \, \text{ for every $w \in W$}$$

Proposition: With the notation as above, $\psi(A,B)=h_W\big(\tr_{V} (A \otimes B)\big)$ where $h_W:W \otimes \Lambda_2(W) \to W$ is defined by the linear extension of

$$ \tilde w \otimes (w_1 \wedge w_2) \to \langle \tilde w,w_2 \rangle w_1-\langle \tilde w,w_1 \rangle w_2. \tag{6}$$

Note $A \otimes B \in V^* \otimes V^* \otimes W \otimes \Lambda_2(W)$, so $\tr_{V} (A \otimes B) \in W \otimes \Lambda_2(W)$.

Proof:

It suffices to prove this for $A,B$ "pure" tensors, i.e $A=\alpha \otimes \tilde w,B=\beta \otimes (w_1 \wedge w_2)$, where $\alpha,\beta \in V^*,\tilde w,w_1,w_2 \in W$.

Now, on the one hand

$$ \langle w \wedge A,B \rangle_{\Hom(V,\Lambda_2(W))}= \langle \alpha \otimes (w \wedge \tilde w) ,\beta \otimes (w_1 \wedge w_2) \rangle_{\Hom(V,\Lambda_2(W))}=$$

$$ \langle \alpha , \beta \rangle_{V^*} \langle w \wedge \tilde w ,w_1 \wedge w_2 \rangle_{\Lambda_2(W)}. $$

On the other hand

$$ \tr_{V} (A \otimes B)= \langle \alpha , \beta \rangle_{V^*} \tilde w \otimes (w_1 \wedge w_2).$$

Thus, it's enough to show

$$ \langle w \wedge \tilde w ,w_1 \wedge w_2 \rangle_{\Lambda_2(W)}=\langle w , h_W\big(\tilde w \otimes (w_1 \wedge w_2)\big) \rangle_W,$$

but this nows follows directly form the definition of the induced inner product on $\Lambda_2(W)$, and the definition of $h_W$ (see $(6)$).

Using the above proposition, we deduce that $$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)\big)\bigg).$$

Can we find a more nice expression for $h_{\phi^*TN}$ ($h_W$)?


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Asaf Shachar
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Asaf Shachar
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