Automatic transfer of pointwise metric computations to bundle computations

Question

$\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$ $\newcommand{\deriv}[2]{\frac{d#1}{d#2}}$ $\newcommand{\sAverage}[1]{\langle#1\rangle} $ $\newcommand{\IP}[2]{\sAverage{#1,#2}}$ $\newcommand{\Cof}{\text{Cof}}$ $\newcommand{\Hom}{\operatorname{Hom}}$ $\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\TM}{\operatorname{T\M}}$ $\newcommand{\TN}{\operatorname{T\N}}$ $\newcommand{\TstarM}{\operatorname{T^*\M}}$

There is a well-known folklore saying that "any linear algebraic construction/statement can be lifted to vector bundles" (e.g tensor products, direct sums, quotients etc).

I am interested in a metric version of this phenomena:

Does every statement about inner-prodcut spaces admit a vector bundle analog?

Specifically, I am interested in "derivations-type" results:

On various occasions, I need to compute derivatives of certain "geometric quantities" associated with bundle maps over a manifold. (examples are given below).

Often, I find it's easier to start with a finite dimensional analogous computation. The computation in the bundle context then becomes a routine adaptation of the original calculation, modulu some extra justifications (revolving around the compatiblity of connections with metrics).

Soft Question: Is there a way to "automate" this transfer? (I want to avoid repeating essentially the same calculation twice). In other words, is there a way to prove a "meta-theorem" which says that the result in the pointwise context carries over to the bundle context?

Main Example: Calculating the derivative of the determinant.

We want to prove the following:

Theorem 1: Let $f:\M \to \N$ be a smooth map between $d$-dimensional oriented Riemannian manifolds. Define $\Cof df= (-1)^{d-1} \star_{f^*TN}^{d-1} (\wedge^{d-1} df) \star_{TM}^1.$ Then for all $V \in \Gamma(\TM)$ $$ d(\det df)(V)= \IP{\Cof df}{\nabla_V df}_{\TM,f^*{\TN}} . $$

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We start first with the "pointwise" analogy, where the vector spaces are fixed and only the linear map is changing:

Proposition: (The cofactor is the gradient of the determinant) Let $V,W$ be oriented $d$-dimensional inner product spaces. Then $$d(\det)_A(B)=\tr\left( \Cof A^T B \right)=\IP{\Cof A}{B}_{V,W}.$$

Specific question: Can we deduce the theorem from the proposition? (without using the proof of the proposition, like I am doing below).

One obvious way to achieve this would be to view $p \to \det(df_p)$ as the determinant of a changing map between fixed vector spaces. This can be done by representing $df$ w.r.t orthonormal frames. However, one then needs to track the derivative of this matrix in terms of $V$ which looks cumbersome. (I would say that even if this approach would work, it is less aesthetic - an invariant way would be better).

Edit:

As pointed out by Deane Yang, there is a more general version of theorem $1$ which is the right "bundle-analog" of the finite-dim proposition:

Theorem 2: Let $E$ and $F$ be rank $d$ oriented vector bundles over $\M$ with smooth metrics and compatible connections. Let $A:E \to F$ be a smooth bundle map. Define $\Cof A= (-1)^{d-1} \star_{F}^{d-1} (\wedge^{d-1} A) \star_{E}^1.$ Then for all $V \in \Gamma(\TM)$ $$ d(\det A)(V)= \IP{\Cof A}{\nabla_V A}_{E,F}. $$

The proof of theorem $2$ is exactly the same as the proof of theorem 1 (see below) - we just replace $df \to A$ everywhere (that proof does not use the fact $df$ is the differential of a map, just the bundle-structures).

The question still remains- can we use the statement of the proposition to deduce theorem $2$, without looking at the proof. (This is not a trivial consequence of the proposition, where the two vector spaces, while different, are fixed).

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proof of the proposition: Let $A_t$ be a smooth family of mappings in $\Hom(V,W)$: $A(0)=A,A'(0)=B$, and let $e_1,\dots,e_d$ be a positive orthonormal basis of $V$. $$ \det(A_t)= \star^d_W \circ \bigwedge^d A_t \circ \star^0_V(1)= \star^d_W \bigwedge^d A_t \big( e_1 \wedge \dots \wedge e_d \big)= \star^d_W \big( A_t e_1 \wedge \dots \wedge A_te_d \big) $$

Using the Leibniz rule we get: $$ \left. \deriv{\det A_t}{t} \right|_{t=0}= \star^d_W \left. \deriv{}{t} \right|_{t=0}\big( A_t e_1 \wedge \dots \wedge A_te_d \big) =$$ $$ \star^d_W \sum_{i=1}^d \big( A e_1 \wedge \dots \wedge Be_i \wedge \dots \wedge Ae_d \big) = \sum_{i=1}^d \star^d_W (-1)^{i-1} \big( Be_i \wedge A e_1 \wedge \dots \wedge \widehat Ae_i \wedge \dots \wedge Ae_d \big) $$ $$ =\sum_{i=1}^d \star^d_W (-1)^{i-1} \big( Be_i \wedge (-1)^{d-1} \star^1_W \star^{d-1}_W ( A e_1 \wedge \dots \wedge \widehat Ae_i \wedge \dots \wedge Ae_d) \big) = $$ $$(-1)^{i-1} (-1)^{d-1} \sum_{i=1}^d \star^d_W \big( Be_i \wedge \star^1_W \star^{d-1}_W ( A e_1 \wedge \dots \wedge \widehat Ae_i \wedge \dots \wedge Ae_d) \big) = $$ $$(-1)^{i-1} (-1)^{d-1} \sum_{i=1}^d \IP{Be_i}{ \star^{d-1}_W ( A e_1 \wedge \dots \wedge \widehat Ae_i \wedge \dots \wedge Ae_d)}_W = $$ $$ (-1)^{d-1} \sum_{i=1}^d \IP{Be_i}{ \star^{d-1}_W \big( \bigwedge^{d-1} A ( \star_V^1 e_i )\big)}_W=\sum_{i=1}^d \IP{Be_i}{ \big( (-1)^{d-1} \star^{d-1}_W \bigwedge^{d-1} A \star_V^1\big) e_i }_W=$$ $$\sum_{i=1}^d \IP{Be_i}{ \Cof A(e_i) }_W=\sum_{i=1}^d \IP{(\Cof A)^TBe_i}{ e_i }_V=\tr\left( \Cof A^TB \right)= \IP{\Cof A}{B}_{V,W}.$$

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proof of Theorem $1$: We want to imitate the proof above:

A positive orthonormal frame $e_1,\dots,e_d$ of $\TM$ will replace the basis for $V$, covariant differentiation will replace the time derivation, so $A \to df,A'(0)=B \to \nabla_Vdf$. There are two obstacles with using this analogy verbatim:

1. The $e_i$ cannot be chosen to be parallel w.r.t $\nabla^{\TM}$ if $\M$ is not flat, while in the original setting, the $e_i$ were constant vectors (time-independent).
2. The derivative w.r.t time commuted with $\star_W$, since it was a fixed linear operator. This time we need to establish a stronger commutation property between Hodge duals and covariant differentiation.

We shall see that a miracle will happen - metricity shall come to our aid. $$ \det(df)= \star^d_{f^*T\N} \circ \bigwedge^d df \circ \star^0_{\TM}(1)= \star^d_{f^*T\N} \big( df(e_1) \wedge \dots \wedge df(e_d) \big),$$ So $$ V\det df = V \star^d_{f^*T\N}\big( df(e_1) \wedge \dots \wedge df(e_d) \big) \stackrel{(1)}{=} $$ $$ \star^d_{f^*T\N} \nabla_V \big( df(e_1) \wedge \dots \wedge df(e_d) \big)= \star^d_{f^*T\N} \sum_{i=1}^d \big( df(e_1) \wedge \dots \wedge \nabla_V \big(df(e_i)\big) \wedge \dots \wedge df(e_d) \big) = \star^d_{f^*T\N} \sum_{i=1}^d \big( df(e_1) \wedge \dots \wedge (\nabla_V df)e_i \wedge \dots \wedge df(e_d) \big) + $$ $$ \star^d_{f^*T\N} \sum_{i=1}^d \big( df(e_1) \wedge \dots \wedge df(\nabla_{V}e_i) \wedge \dots \wedge df(e_d) \big) \stackrel{(2)}{=} $$ $$ \IP{\Cof df}{\nabla_V df}_{\TM,f^*{\TN}}+ \star^d_{f^*T\N} \bigwedge^d df( \sum_{i=1}^d e_1 \wedge \dots \wedge \nabla_Ve_i \wedge \dots \wedge e_d)= \IP{\Cof df}{\nabla_V df}_{\TM,f^*{\TN}}+ \star^d_{f^*T\N} \bigwedge^d df\big( \nabla_V (e_1 \wedge \dots \wedge e_i \wedge \dots \wedge e_d) \big)=\IP{\Cof df}{\nabla_V df}_{\TM,f^*{\TN}}. $$

Where equality $(1)$ follows since metric connections and Hodge duals commute, and equality $(2)$ is exactly the formal repeatition of the calculation in the pointwise setting (where $A \to df,B \to \nabla_Vdf$).

Admittedly, this repeatition is not huge, but I have other examples on my mind where the computations are much longer, so a general "transfer-principle" would be nice to have.

Answer 1

The following is the meta-theorem I have in mind (I can't swear that what I've written is 100% correct):

Given $G < \mathrm{GL}(n)$, let $\Phi: \mathbb{R}^n \rightarrow \mathbb{R}$ a smooth $G$-invariant function. Let $\Phi': \mathbb{R}^n \rightarrow (\mathbb{R}^n)^*$ denote the differential of $\Phi$, defined to be $$ \langle\Phi'(x),v\rangle = \left.\frac{d}{dt}\right|_{t=0}\Phi(x + tv). $$ Let $V$ be a rank $n$ vector bundle with structure group $G$ (using the same representation as above) and compatible connection $\nabla$. The function $\Phi$ induces naturally a function $\widehat\Phi: V \rightarrow \mathbb{R}$, which is equal to $\Phi$ on each fiber $V_x$. Similarly, $\Phi'$ induces a bundle map $\widehat\Phi': V \rightarrow V^*$.

Given any section $v$ of $V$, let $f(x) = \widehat\Phi(v(x))$. Then given any tangent vector $t \in T_x\mathcal{M}$, $$ \langle df(x), t\rangle = \langle \widehat\Phi'(v(x)), \nabla_tv(x)\rangle. $$

I think this (or something like it) can be proved by using a curve passing through $x$ and tangent to $t$ and a $G$-frame parallel translated along the curve and writing the section $v$ in terms of the frame.

Answer 2

I think everything can be reduced to the following (stated without proofs):

Let $E$ and $F$ be rank $n$ oriented vector bundles over $\mathcal{M}$ with smooth inner products and compatible connections. Let $\omega$ be the section of $\Lambda^nE$ such that $$ \omega = e_1\wedge\cdots\wedge e_n $$ for any local positively oriented orthonormal frame $e_1, \cdots, e_n$ of $E$. Define $\eta \in \Gamma(\Lambda^nF)$ similarly. Compatibility of the connection with the inner product on each bundle implies that $\nabla \omega = 0$ and $\nabla\eta = 0$.

A smooth bundle map $A: E\rightarrow F$ naturally induces a map $A: \Lambda^nE \rightarrow \Lambda^nF$. Let $\det A$ be the real-valued function such that $$ A(\omega) = (\det A)\eta. $$ Define the cofactor of $A$ to be a bundle map $A^c: F\rightarrow E$, which agrees with the standard definition of the cofactor matrix, when $A$ is written with respect to local orthonormal frames of $E$ and $F$. In particular, $A^cA = (\det A)1_E$.

Then, given any smooth vector field $V$ on $\mathcal{M}$, $$ \langle d(\det A),V\rangle = \tr A^c\nabla_VA, $$ where $\nabla$ is the connection on $F\otimes E^*$ induced by the connections on $E$ and $F$. In general, anything invariant formula involving derivatives of maps or tensors of finite dimensional vector spaces translates into a corresponding formula involving vector bundles, where differentiation is replaced by covariant differentiation. A compatible inner product is needed on the bundles, if the vector space formula relies on inner products, too.

In your example, I believe this gives the right formula when $E = T_*\mathcal{M}$ and $F=f^*T_*\mathcal{N}$. Note that the connection on $F$ is the pullback of the Levi-Civita connection on $N$, so, if $x^1, \dots, x^d$ are local coordinates on $\mathcal{M}$ and $y^1, \dots, y^d$ on $\mathcal{N}$, then $$ \nabla_{X_i}Y_\alpha = \frac{\partial f^\beta}{\partial x^i}\Gamma^\gamma_{\alpha\beta}Y_\gamma, $$ where $$ X_i = \frac{\partial}{\partial x^i},\ Y_\alpha = \frac{\partial}{\partial y^\alpha}, $$ and $\Gamma^\gamma_{\alpha\beta}$ are the Christoffel symbols for the metric on $Y$.

Last quick personal comment: I find the Hodge star operator really confusing to work with and often try to avoid it.

Answer 3

$\newcommand{\M}{\mathcal{M}}$ $\newcommand{\Hom}{\operatorname{Hom}}$ $\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\TM}{\operatorname{T\M}}$ $\newcommand{\sAverage}[1]{\langle#1\rangle} $ $\newcommand{\IP}[2]{\sAverage{#1,#2}}$ $\newcommand{\Cof}{\text{Cof}}$

Here is a sort of "meta-theorem:

Background:

Let $V,W$ be oriented $d$-dimensional inner product spaces, and let $\phi:\Hom(V,W) \to \mathbb{R}$ a smooth function which is "defined canonically" using only the structures of metric and orientation. We call such a function $\phi$ a geometric function.

(This notion needs to be made more precise).

Examples: $A \to \det A,A \to \tr A,A \to \text{ the i-th coefficient of the characteristic polynomial of $A$}$

Recall the gradient of $\phi$ is defined by requiring $$d\phi_A(B)=\IP{\nabla \phi(A)}{B}_{V,W}. $$

Note $\nabla \phi$ can be viewed as a map $\Hom(V,W) \to \Hom(V,W)$, and this is also "canonically defined".

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Statement:

Let $E$ and $F$ be rank $d$ oriented vector bundles over $\M$ with smooth metrics and compatible connections.

Let $\phi$ be a geometric function. Then $\phi$ induces a map $$ \tilde \phi :\Hom(E,F) \to C^{\infty}(\M),$$ by acting pointwise.

Similarly, $\nabla \phi$ induces a map:

$$ \widetilde{\nabla \phi} :\Hom(E,F) \to \Hom(E,F).$$

Then, for all $V \in \Gamma(\TM),A \in \Hom(E,F)$

$$ V\cdot \tilde \phi(A)=\IP{\widetilde{\nabla \phi}(A)}{\nabla^{\Hom(E,F)}_VA}_{E,F}.$$

I am not sure how to prove this. (I tried using the chain rule and pullback connection along a path but failed. I think the definition of goemetric function should be made precise in order to establish the proof).

Further comments:

1. In the example of $\phi(A)=\det A$, we recover theorem $2$ from the question, since $ \nabla \phi(A)=\Cof A$.
2. This metha-theorem can probably be generalized to geometric maps, that is maps from $\Hom(V,W)$ which takes values in spaces "constructed from $V,W$ and $\mathbb{R}$" (not only $\mathbb{R}$). A natural exmaple is the cofactor itself: $A \to \Cof A$ is a map $\Hom(V,W) \to \Hom(V,W)$