Exists homeomorphism which carries each fiber isomorphically to itself, composition?

Question

Let $\mu$ and $\mu'$ be two different Euclidean metrics on the same vector bundle $\xi$. How do I see that there exists a homeomorphism $f: E(\xi) \to E(\xi)$ which carries each fiber isomorphically onto itself, so that the composition $\mu \circ f: E(\xi) \to \mathbb{R}$ is equal to $\mu'$?

Thoughts so far. Every positive definite matrix $A$ can be expressed uniquely as the square of a positive definite matrix $\sqrt{A}$. The power series expansion$$\sqrt{tI + X} = \sqrt{t}\left( I + {1\over{2t}} - {1\over{8t^2}}X^2 + \dots\right),$$is valid provided that the characteristic roots of $tI + X = A$ lie between $0$ and $2t$. This shows that the function $A \mapsto \sqrt{A}$ is smooth. But I am not sure how to complete. Can anyone help?

Notation. Let $B$ denote a fixed topological space, which will be called the base space. A real vector bundle $\xi$ over $B$ consists of the following:

• a topological space $E = E(\xi)$ called the total space,
• a (continuous) map $\pi: E \to B$ called the projection map, and
• for each $b \in B$ the structure of a vector space over the real numbers in the set $\pi^{-1}(b)$.

A Euclidean vector bundle is a real vector bundle $\xi$ together with a continuous function$$\mu: E(\xi) \to \mathbb{R}$$such that the restriction of $\mu$ to each fiber of $\xi$ is positive definite and quadratic. The function $\mu$ itself will be called a Euclidean metric on the vector bundle $\xi$.

Answer 1

I'm going to talk about this in terms of inner products instead of quadratic forms, since I'm more comfortable with them. Suppose I have an inner product space $V$. How do I construct an isometry with $\Bbb R^n$ with the standard inner product? Gram-Schmidt: I pick any linear isomorphism to $\Bbb R^n$ and renormalize it to make it into an isometry.

Now suppose I want to do the same thing between inner product spaces $V \to V'$. I no longer have the fortune of having a basis auto-picked for me, so I start by picking an orthonormal basis of $V$, an isomorphism $V \to V'$, and now I can do Gram-Schmidt. One verifies that the resulting linear map did not depend on the choice of orthonormal basis we started with.

I'm going to do the exact same thing on the level of vector bundles. I have a rather obvious isomorphism $E(\xi) \to E(\xi)$ - the identity! Now I'm going to do the Gram-Schmidt process fiberwise. I started with isomorphisms $\xi_p \to \xi_p$ and I ended with isometries $f_p: (\xi_p, \mu) \to (\xi_p, \mu')$.

What we need is for the Gram-Schmidt process to be continuous. This means that if I have two pairs of metrics $\mu_i$ on $V$ and $\mu_i'$ on $V'$ that are 'nearby', and a specified linear map between them, then the maps Gram-Schidt gives me for $(\mu_i, \mu_i')$ are 'nearby' as well. (To be careful, I would be using $\varepsilon$s and $\delta$s.) This isn't anything particularly clever: it's just that $\text{proj}_u v$ will be close, and so will $u/\|u\|$. I don't really want to write out the details, but they are straightforward.

It's not clear to me where the $\sqrt{A}$ thing shows up. I guess if I had phrased everything in terms of quadratic forms it would be there in whatever Gram-Schmidt becomes instead (so I'm checking continuity of $\sqrt{A}$ as opposed to continuity of Gram-Schmidt.)

Answer 2

There are several questions asking about the same problem on stackexchange: Different Euclidean metrics on a vector bundle, Isometry of two Euclidean structures on the same vector bundle, but I haven't seen a complete correct answer. This answer https://math.stackexchange.com/a/1209444/251687 is correct but lacks some details; the others I saw were wrong.

The question is Problem 2-E in Milnor's book Characteristic Classes, and I think the author's hint is crucial, which I list as the following lemma.

Lemma. For any positive definite matrix $A$, there exists a unique positive definite matrix $B$ such that $B^2=A$. If we write $\sqrt{A}:=B$, then the map $A\mapsto\sqrt{A}$ is smooth.

Equivalently, for a positive definite matrix $A$, $\sqrt{A}$ is the unique positive definite matrix such that $\sqrt{A}^t\sqrt{A}=A$.

Now we can give two proofs of the problem. The first one is in terms of local trivializations; the second one is formulated in an intrinsic manner.

Proof. (Define isometries locally and check they agree on overlaps.)

By choosing local $\mu$-orthonormal basis vectors, we can cover the base manifold $M$ by $\{U_\alpha\}$ such that there are local isometric trivializations $(E|_{U_\alpha},\mu)\xrightarrow{\phi_\alpha}(U_\alpha\times\mathbb{R}^n,\mu_{\mathbb{R}^n})$, where $\mu_{\mathbb{R}^n}$ is standard metric. Let $g_{\alpha\beta}:U_\alpha\cap U_\beta\to O(n)$ be the transition maps. In each local trivialization, the inner product $((\phi_\alpha^{-1})^*\mu^\prime)|_{U_\alpha\times\mathbb{R}^n}=:\mu^\prime_{\mathbb{R}^n_\alpha}$ is given by a positive definite matrix-valued function $A_\alpha$, and $A_\alpha=g_{\alpha\beta}A_\beta g_{\alpha\beta}^{-1}$. Now define $\psi_\alpha:U_\alpha\times\mathbb{R}^n\to U_\alpha\times\mathbb{R}^n$ by $\psi_\alpha(x,v)=(x,\sqrt{A_\alpha(x)}v)$, then for any two vectors $v,w\in \mathbb{R}^n=\phi_\alpha(E_x)$, $(\psi_\alpha^*\mu_{\mathbb{R}^n})(v,w)=v^t\sqrt{A_\alpha}^t\sqrt{A_\alpha}w=v^tA_\alpha w=\mu^\prime_{\mathbb{R}^n_\alpha}(v,w)$, and hence $\mu^\prime|_{E|_{U_\alpha}}=(\phi_\alpha^{-1}\psi_\alpha\phi_\alpha)^*(\mu|_{E|_{U_\alpha}})$. So $\varphi_\alpha:=\phi_\alpha^{-1}\psi_\alpha\phi_\alpha$ gives an isometry $(E|_{U_\alpha},\mu^\prime)\to(E|_{U_\alpha},\mu).$ Next we want to check such $\varphi_\alpha=\varphi_\beta$ on overlap. (Here the notation is different from what I meant in my comment on Mike's answer.) This means $\phi_\alpha^{-1}\psi_\alpha\phi_\alpha=\phi_\beta^{-1}\psi_\beta\phi_\beta$, i.e. $\psi_\alpha=g_{\alpha\beta}\psi_\beta g_{\alpha\beta}^{-1}$. By definition of $\psi_\alpha$, what we want is $\sqrt{A_\alpha}=g_{\alpha\beta}\sqrt{A_\beta} g_{\alpha\beta}^{-1}$. But $\sqrt{A_\alpha}$ and $g_{\alpha\beta}\sqrt{A_\beta} g_{\alpha\beta}^{-1}$ are both positive definite matrices whose square equals $A_\alpha$, so by uniqueness, we are done.

In the same spirit, we can give a more intrinsic proof.

Alternative Proof. (More intrinsic.)

We can rephrase the lemma as: Given any finite dimensional real vector space $V$ and two inner products $\mu,\mu^\prime$ on $V$, there exists a unique $\varphi\in GL(V)$, depending smoothly on $\mu,\mu^\prime$, such that

• $\varphi$ is self-adjoint w.r.t. $\mu$;

• $\varphi:(V,\mu^\prime)\to(V,\mu)$ is an isometry.

So now for our vector bundle $E$, one can define a unique isometry $\varphi_x:(E_x,\mu^\prime_x)\to(E_x,\mu_x)$ on each fiber $E_x$ s.t. $\varphi_x$ is self-adjoint w.r.t. $\mu_x$. This gives a global smooth isometry $\varphi:(E,\mu^\prime)\to(E,\mu)$.

Remark. The same conclusion holds true in the setting of Hermitian metrics on complex vector bundles. In other words, given a vector bundle $E\to M$, the general linear gauge group $GL(E)$ acts transitively on the space of Hermitian/Euclidean metrics on $E$. So one can study the orbit space of such action, which is $GL(E)/U(E,h_0)$ where $h_0$ is a prescribed metric. This is one way of thinking about Hermitian-Yang-Mills connections on stable holomorphic bundles.