Is the induced connection in the hom bundle compatible with the metric provided that the connections from which it comes are?

Question

$\def\bbR{\mathbb{R}} \def\frX{\mathfrak{X}} \def\hb{\mathcal{H}om} \def\hom{\operatorname{Hom}}$Let $E$ be a smooth vector bundle over a smooth manifold $M$. A metric in $E$ is a bundle homomorphism $E\otimes E\to M\times\bbR$ which is fiberwise positive-definite. If $E$ has a metric, a connection $\nabla$ on $E$ is said to be metric or compatible with the metric if $$ \nabla_X\langle Y,Z\rangle=\langle\nabla_XY,Z\rangle+\langle Y,\nabla_XZ\rangle $$ for all $Y,Z\in\Gamma(E)$ and $X\in\frX(M):=\Gamma(TM)$. Now, suppose $F$ is another vector bundle over $M$. Denote $\hb(E,F)$ to the hom-bundle of $E$ and $F$, i.e., it is the vector bundle over $M$ whose fiber over $p\in M$ is $\hb(E,F)_p=\hom(E_p,F_p)$. The connections on $E$ and on $F$ induce a connection on the hom-bundle in the following way: for $\varphi\in\Gamma(\hb(E,F))=\hom(E,F)$, we have $$ (\nabla_X\varphi)Y=\nabla_X(\varphi(Y))-\varphi(\nabla_XY), $$ for $X\in\frX(M),Y\in\Gamma(E)$ (details here).

Now suppose that $F$ has a metric. The metrics of $E$ and $F$ induce then a metric on $\hb(E,F)$ which fiberwise is given as here.

I was wondering if the connections on $E$ and $F$ being metric implies that the induced connection on $\hb(E,F)$ is metric as well. Here are my thoughts on this: Pick an orthonormal frame $(E_i)$ of $E$ (we can achieve this by picking first a local frame and then applying Gram-Schmidt. Since the GS formulas are smooth, so is the resulting frame).

Let $\varphi,\psi\in\Gamma(\hb(E,F))$, $X\in\frX(M)$, and compute \begin{align*} \nabla_X\langle\varphi,\psi\rangle &=\sum_i\nabla_X\langle\varphi(E_i),\psi(E_i)\rangle\\ &=\sum_i\langle\nabla_X(\varphi E_i),\psi E_i\rangle +\langle\varphi E_i,\nabla_X(\psi E_i)\rangle\\ &=\sum_i\langle(\nabla_X\varphi)E_i+\varphi(\nabla_X E_i),\psi E_i\rangle +\langle \varphi E_i,(\nabla_X\psi)E_i+\psi(\nabla_X E_i)\rangle\\ &=\sum_i\langle(\nabla_X\varphi)E_i,\psi E_i\rangle+\langle\varphi E_i,(\nabla_X\psi)E_i\rangle\\ &\quad +\sum_i\langle\varphi(\nabla_X E_i),\psi E_i\rangle+\langle\varphi E_i,\psi(\nabla_X E_i)\rangle. \end{align*} The second-to-last line equals $\langle\nabla_X\varphi,\psi\rangle+\langle\varphi,\nabla_X\psi\rangle$, but what about the last line? Does this expression vanish? In case it did, it would suffice to show it for $X=E_j$, by $C^\infty(M)$-linearity in $X$. In this case the last line becomes $$ \sum_i\langle\varphi(\nabla_X E_i),\psi E_i\rangle+\langle\varphi E_i,\psi(\nabla_X E_i)\rangle =\sum_i\Gamma_{ji}^k\left(\langle\varphi E_k,\psi E_i\rangle+\langle\varphi E_i,\psi E_k\rangle\right). $$ But I still regard this a little intractable.

Answer 1

The crucial observation is that there is a natural isomorphism $$\operatorname{Hom}(E,F) \simeq F\otimes E^*.$$ Also, it suffices to define and verify everything for simple tensors $f\otimes \epsilon \in F\otimes E^*$.

If $E$ has an inner product, then there is a naturally induced metric on $E^*$. If $F$ also has an metric, then there is e naturally induced metric on $F\otimes E^*$, where for any $f \in F$ and $\epsilon\in E^*$, $$ \langle f_1\otimes \epsilon_1,f_2\otimes\epsilon_2\rangle = \langle f_1,f_2\rangle\langle \epsilon_1,\epsilon_2\rangle. $$

If $E$ has a connection, then there is a naturally induced connection on $E^*$. If $F$ also has a connection, then there is a naturally induced connection on $F\otimes E^*$, where if $f$ is a section of $F$, $\epsilon$ is a section of $E^*$, and $X \in TM$, $$ \nabla_X(f\otimes\epsilon) = (\nabla_X f)\otimes\epsilon + f\otimes \nabla_X\epsilon. $$

It is now a straightforward calculation to show that if the connections on $E$ and $F$ are compatible with the respective metrics on $E$ and $F$, then, given sections $f_1, f_2$ of $F$ and $\epsilon_1,\epsilon_2$ of $E^*$, and $X \in TM$, \begin{align*} X\langle f_1\otimes \epsilon_1, f_2\otimes \epsilon_2\rangle &= X(\langle f_1,f_2\rangle\langle \epsilon_1,\epsilon_2\rangle)\\ &= \cdots\\ &= \langle\nabla_X(f_1\otimes\epsilon_1),f_2\otimes\epsilon_2\rangle + \langle f_1\otimes\epsilon_1,\nabla_X(f_2\otimes\epsilon_2)\rangle. \end{align*} In other words, the naturally induced connection on $\operatorname{Hom}(E,F)$ is compatible with the naturally induced metric on $\operatorname{Hom}(E,F)$.

Answer 2

The short answer is yes.

The easiest long answer I can think of uses the language of principal bundles and associated vector bundles. If you have not encountered these objects, then I can recommend a few sources for you, but the basics of them isn't too hard to wrap ones head around.

Let $E$ and $F$ be smooth vector bundles over $M$ with covariant derivatives $\nabla^E$ and $\nabla^F$ which are metric compatible with the bundle metrics $\langle \cdot ,\cdot \rangle_E$ and $\langle \cdot, \cdot \rangle_F$. Note that the induced bundle metric on $E\otimes F$ is given pointwise on simple tensors by :

$$\langle \phi_E\otimes \phi_F,\psi_E\otimes \psi_F \rangle_{(E\otimes F)_x}=\langle \phi_E ,\psi_E \rangle_{E_x}\cdot \langle \phi_F,\psi_F \rangle_{F_x}$$

First note that since $E$ and $F$ metrics on them, there exists a bundle of orthonormal frames of $E$ and $F$, which we denote by $O(E)$, and $O(F)$ respectively. If the fibre of $E$ is isomorphic to $\mathbb{R}^n$, and the fibre of $F$ is isomorphic to $\mathbb{R}^m$ we have that $O(E)$ and $O(F)$ admit smooth right actions of $O(n)$ and $O(m)$ respectively, which preserve the fibres of the bundles, and are free and transitive on them.

We can then describe $E$ and $F$ as vector bundles associated these principal bundles. Indeed, if $\rho_n$ and $\rho_m$ denote the standard representations of $O(n)$ and $O(m)$ on $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively, then we have that as vector bundles:

$$E\cong O(E)\times_{\rho_n}\mathbb{R}^n$$

where $O(E)\times_{\rho_n}\mathbb{R}^n$ is the quotient space defined by the equivalence relation: $$(p,v)\sim (p\cdot g, \rho_n(g)^{-1} v)$$ Writing points in $O(E)$ as a frame $(v_1,\dots, v_n)_p$ for $E_p$, and a vector in $\mathbb{R}^n$ as $(x^1,\dots, x^n)$, the isomorphism is given by:

$$[(v_1,\dots, v_n)_p, (x^1,\cdots, x^n)]\longmapsto v_ix^i$$

It is easy to see that this assignment is independent of our choice of class representative, and checking that this is indeed smooth and a fibre wise vector space isomorphism is also done easily. Similar results hold for $F$.

Let $\langle \cdot ,\cdot \rangle_{\mathbb{R}^n}$ be the standard inner product on $\mathbb{R}^n$, then the original bundle metric $E$ can be rewritten as: $$\langle [p,v], [p,w]\rangle_{E_x}= \langle v,w\rangle_{\mathbb{R}^n}$$ where $\pi(p)=x$. This should make intuitive sense, as this operation is well defined since $\rho_n$ leaves $\langle \cdot ,\cdot \rangle_{\mathbb{R}^n}$ invariant so this is well defined, and in this framework we are essentially writing everything in terms of orthonormal frames of $E$.

The metric compatible covariant derivative's $\nabla^E$, $\nabla^F$ induce connection one forms $A_E$ and $A_F$ in the principal bundles $O(E)$ and $O(F)$. These are one forms valued in the Lie algebras $\mathfrak{o}(n)$ and $\mathfrak{o}(m)$, and satisfy two properties. Namely, that under pullback by right multiplication we have: \begin{align*} R_g^* A_E=Ad_{g^{-1}}\circ A_E \end{align*} and for all fundamental vector field $\tilde{X}$ associated to a Lie algebra element $X$ we have: \begin{align*} A_E(\tilde{X})=X \end{align*} Here, $\tilde{X}$ is defined pointwise by: \begin{align*} \tilde{X}_p=\frac{d}{dt}\Big|_{t=0}p\cdot \exp(tX) \end{align*} Furthermore, we can reobtain the metric compatible covariant derivative by defining a covariant derivative locally by noting that for a smooth section $s:U\rightarrow O(E)_U$, local sections $\Phi:U\rightarrow E_U$, are in one to correspondence with smooth maps $\phi:U\rightarrow \mathbb{R}^n$. It follows that given a global section $\Phi:M\rightarrow E$, and a local section $s:U\rightarrow O(E)_U$:

$$\Phi|_U=[s, \phi]$$

for some unique smooth map $\phi:U\rightarrow \mathbb{R}^n$. We thus define the covariant derivative locally by:

$$\nabla^E_X\Phi|_U=[s, d\phi(X)+\rho_{n*}(A_{E_s}(X))\phi]$$

where $\rho_{n*}$ is the induces representation of $\mathfrak{o}(n)$ on $\mathbb{R}^n$, and $A_{E_s}$ is the pull back of $A_E$ by $s$. One can check that this is independent of the section we chose, so we this defines a global operation. Furthermore, since: \begin{align} \langle \rho_{n*}(X)v, w\rangle_{\mathbb{R}^n}+\langle v, \rho_*(X)w \rangle_{\mathbb{R}^n}=0 \end{align} for all $X\in \mathfrak{o}(n)$, one can check that this prescription is metric compatible with the bundle metric $\langle \cdot ,\cdot \rangle_E$. We can do similar things for $F$.

Ok, but what does this tell us about $E\otimes F$? Well it turns out we can also describe $E\otimes F$ as vector bundle associated to some principal bundle: the fibre wise product of $O(E)$ and $O(F)$. Indeed, we define the fibrewise product by:

$$O(E)\times_M O(F)=\coprod_{x\in M} \pi_{O(E)}^{-1}(x)\times \pi_{O(F)}^{-1}(x)$$

As a set, this is equivalent to:

$$\{(p,q)\in O(E)\times O(F): \pi_{O(E)}(p)=\pi_{O(F)}(q)\}$$

This set can be shown to be principal $O(n)\times O(m)$ principal bundle, where for all $(g,h)\in O(n)\times O(m)$, the right action is given by:

$$(p,q)\cdot (g,h)=(p\cdot g,q\cdot h)$$

Furthermore, we obtain a new representation $\rho_n\otimes \rho_m$ on $\mathbb{R}^n\otimes \mathbb{R}^m$ given on simple tensors by:

$$(\rho_n\otimes \rho_m)(g,h)\cdot(v\otimes w)=\rho_n\cdot(g)v\otimes \rho_m(h)\cdot w$$

We then have that as vector bundles:

$$E\otimes F\cong (O(E)\times _M O(F))\times_{\rho_n\otimes \rho_m} (\mathbb{R}^n\otimes \mathbb{R}^m)$$ We have an induced inner product on $\langle\cdot,\cdot \rangle_{\mathbb{R}^n\otimes \mathbb{R}^m}$, given by the standard inner products on $\mathbb{R}^n$ and $\mathbb{R}^M$. Via the same process as before, this induces the the inner product on $E\otimes F$.

The connection one forms on $O(E)$ and $O(F)$ define a new connection one form $A_E\oplus A_F$ on $O(E)\times_M O(F)$ given by:

$$(A_E\oplus A_F)_(p,q)(X_p,Y_q)=(A_E(X_p), A_F(Y_q))$$

for all $(X_p,Y_q)\in T_pO(E)\times T_qO(F)$ where $\pi_{O(E)}(p)=\pi_{O(F)}(q)$, and $\pi_{O(E)*}X_p=\pi_{O(F)*}Y_q$. This connection one form induces the correct covariant derivative on $E\otimes F$. Let's see this; let $\Phi\in \Gamma(E)$ and $\Psi\in\Gamma(F)$, then: \begin{align*} \nabla^{E\otimes F}_X(\Phi\otimes \Psi)|_U=&[s_E\times_M s_F, d(\phi\otimes \psi)+(\rho_{n}\otimes \rho_m)_*((A_{E_{s_E}}\oplus A_{F_{s_F}})(X))(\phi\otimes\psi)]\\ =&[s_E\times s_F,d\phi\otimes \psi+\rho_{n*}(A_{E_{s_E}}(X))\phi\otimes \psi]+[s_E\times s_F,\phi\otimes d\psi+\phi\otimes\rho_{m*}(A_{F_{s_F}}(X))\psi]\\ =&\nabla^E_X\Phi\otimes \Psi|_U+\Phi\otimes \nabla^F_X\Psi|_U \end{align*} We want to check that this is connection is metric compatible, however this trivially follows from the earlier discussion, as the representation $\rho_n\otimes \rho_m$ leaves the inner product $\langle \cdot,\cdot\rangle_{\mathbb{R}^n\otimes \mathbb{R}^m}$ invariant, so we'll obtain a similar rule for the induced representation $(\rho_n\otimes \rho_m)_*$.

So since $Hom(E,F)\cong E^*\otimes F$, we then just need to show that the induced covariant derivative is on $E^*$ is metric compatible. We will use the frame bundle $O(E)$ to construct $E^*$ as a vector bundle associated to $O(E)$. Let $\mathbb{R}^{n*}$ be the dual space to $\mathbb{R}^n$, then there is an induced representation $\rho'_n$ of $O(n)$ on $\mathbb{R}^n$ given implicitly by:

$$(\rho'_n(g)\cdot \omega)(v)=\omega(\rho_n(g^{-1})v)$$

One can then show by the definition of $E^*$:

$$E^*\cong O(E)\times_{\rho'_n} \mathbb{R}^{n*}$$

Note that the inner product on $\mathbb{R}^{n}$ induces an isomorphism $\mathbb{R}^n\rightarrow \mathbb{R}^{n*}$ given by:

$$f:v\longmapsto \langle v,\cdot\rangle_{\mathbb{R}^n}$$ implying that $f(v)(w)=\langle v,\cdot w\rangle$. The inverse $f^{-1}$ induces the inner product on $\mathbb{R}^{n*}$ given by: \begin{align*} \langle \omega, \eta\rangle_{\mathbb{R}^{n*}}=\langle f^{-1}(\omega),f^{-1}(\eta)\rangle_{\mathbb{R}^n} \end{align*} We need to check that $\rho'_n$ leaves this inner product invariant. We have that: \begin{align*} \langle \rho'_n(g)\cdot \omega,\rho'_n(g)\cdot \eta \rangle_{\mathbb{R}^{n*}}=& \langle f^{-1}(\rho'_n(g)\cdot \omega), f^{-1}(\rho'_n(g)\cdot \eta)\rangle_{\mathbb{R}^n} \end{align*} Note that we have: \begin{align*} f(\rho(g)v)(w)=\langle \rho_n(g)v,w\rangle_{\mathbb{R}^n}=\langle v,\rho_n(g)^Tw\rangle_{\mathbb{R}^n} \end{align*} Since $\rho(g)^T=\rho(g)^{-1}$ we obtain that: $$f(\rho_n(g)v)(w)=f(v)(\rho_n(g^{-1})w)=\rho'_n(g)\cdot f(v)$$ It then follows that for some unique $v\in \mathbb{R}^n$: \begin{align*} f^{-1}(\rho'_n(g)\cdot \omega)=&f^{-1}(\rho'_n(g)\cdot f(v))\\ =&f^{-1}(f(\rho_n(g)\cdot v))\\ =&\rho_n(g)\cdot v \end{align*} So if $\omega=f(v)$ , and $\eta=f(w)$ we have that: \begin{align*} \langle \rho'_n(g)\cdot \omega,\rho'_n(g)\cdot \eta \rangle_{\mathbb{R}^{n*}}=& \langle f^{-1}(\rho'_n(g)\cdot \omega), f^{-1}(\rho'_n(g)\cdot \eta)\rangle_{\mathbb{R}^n}\\ =&\langle f^{-1}(f(\rho_n(g)\cdot \omega), f^{-1}(f(\rho_n(g)w))\rangle_{\mathbb{R}^n}\\ =&\langle \rho_n(g)v,\rho_n(g)w\rangle_{\mathbb{R}^n}\\ =&\langle v,w\rangle_{\mathbb{R}^n}\\ =&\langle \omega,\eta\rangle_{\mathbb{R}^{n*}} \end{align*} This inner product then induces the same inner product on $E^*$ as the one obtained by the bundle isomorphism $F:E\rightarrow E^*$ induced by $\langle \cdot ,\cdot \rangle_{E}$. It is easy to check that the covariant derivative defined in the language of connection one forms is then the induced covariant derivative, and since the $\rho'_n$ leaves the inner product on $\mathbb{R}^{n*}$ invariant, it follows that the induced covariant derivative is also metric compatible.

From our earlier discussion we then have that the induced covariant derivative on $E^*\otimes F$ is metric compatible with the induced metric, as desired. You can check that this covariant derivative will be the same as the one you defined implicitly above.