$\newcommand{\M}{\mathcal{M}}$
$\newcommand{\N}{\mathcal{N}}$
$\newcommand{\deriv}[2]{\frac{d#1}{d#2}}$
$\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 $V \in \Gamma(\TM)$
>  $$
>  d(\det df)(V)= \IP{\Cof df}{\nabla\_V df}\_{\TM,f^\*{\TN}} .
>  $$
> 
> 
> 




---


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).




---


*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:
$$
\left. \deriv{\det A\_t}{t} \right|\_{t=0}= \star^d\_W \left. \deriv{}{t} \right|\_{t=0}\big( A\_t e\_1 \wedge \dots \wedge A\_te\_d \big) =$$
$$ \star^d\_W \sum\_{i=1}^d \big( A e\_1 \wedge \dots \wedge Be\_i \wedge \dots \wedge Ae\_d \big) = \sum\_{i=1}^d \star^d\_W (-1)^{i-1} \big( Be\_i \wedge A e\_1 \wedge \dots \wedge \widehat Ae\_i \wedge \dots \wedge Ae\_d \big) $$
$$ =\sum\_{i=1}^d \star^d\_W (-1)^{i-1} \big( Be\_i \wedge (-1)^{d-1} \star^1\_W \star^{d-1}\_W ( A e\_1 \wedge \dots \wedge \widehat Ae\_i \wedge \dots \wedge Ae\_d) \big) = $$
 $$(-1)^{i-1} (-1)^{d-1} \sum\_{i=1}^d \star^d\_W \big( Be\_i \wedge \star^1\_W \star^{d-1}\_W ( A e\_1 \wedge \dots \wedge \widehat Ae\_i \wedge \dots \wedge Ae\_d) \big) = $$
 $$(-1)^{i-1} (-1)^{d-1} \sum\_{i=1}^d \IP{Be\_i}{ \star^{d-1}\_W ( A e\_1 \wedge \dots \wedge \widehat Ae\_i \wedge \dots \wedge Ae\_d)}\_W = $$
 $$ (-1)^{d-1} \sum\_{i=1}^d \IP{Be\_i}{ \star^{d-1}\_W \big( \bigwedge^{d-1} A ( \star\_V^1 e\_i )\big)}\_W=\sum\_{i=1}^d \IP{Be\_i}{ \big( (-1)^{d-1} \star^{d-1}\_W \bigwedge^{d-1} A \star\_V^1\big) e\_i }\_W=$$
 $$\sum\_{i=1}^d \IP{Be\_i}{ \Cof A(e\_i) }\_W=\sum\_{i=1}^d \IP{(\Cof A)^TBe\_i}{ e\_i }\_V=\tr\left( \Cof A^TB \right)= \IP{\Cof A}{B}\_{V,W}.$$




---


*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](http://math.stackexchange.com/questions/1996515/does-hodge-star-commute-with-metric-connections).


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](http://math.stackexchange.com/questions/1996515/does-hodge-star-commute-with-metric-connections), 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.



Automatic transfer of pointwise metric computations to bundle computations
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.


