Isometry of two Euclidean structures on the same vector bundle

I'm reading Milnor & Stasheff's Characteristic Classes, and I was struck by the following problem, which they call the Isometry Theorem:

"Let $\mu$ and $\mu'$ be two different Euclidean metrics on the same vector bundle $\xi$. Prove that there exists a homeomorphism $E(\xi) \rightarrow E(\xi)$ (i.e. of the total space) which carries each fiber isomorphically onto itself such that $\mu \circ f = \mu'$."

The hint for the problem suggests using the square root of a matrix and sketches a proof that the map $A \mapsto \sqrt{A}$ is smooth. I don't see the relevance of square roots anywhere.

My attempt at a proof is to say that between if $M, N$ are two positive definite matrices, then $M = {^{t}A} N A$ for some invertible matrix $A$, and then define the isomorphism of vector bundles on each stalk by $A$. What's wrong with this attempt? Where do square roots come in?

EDIT: Thinking about this question raised another worry: why doesn't this result show that any two Riemannian metrics on the same space are isometric? If I'm remembering correctly, the only extra requirement we need there is that the map depend smoothly on the stalk, which the exercise in Milnor/Stasheff claims is true for square roots.

• There are $O(n)$ many such $A$, and no (obvious) canonical one. You have to pick $A$ in a continuous fashion. Locally, this is fine; but there's no obvious reason you should be able to globally patch together your local choices. The point of the square root business is that it allows you to pick an $A$ canonically, in a fashion that continuously depends on the stalk.
– user98602
Commented Mar 28, 2015 at 0:52
• Regarding your edit: remember that an isometry of Riemannian manifolds is, in particular, a diffeomorphism of Riemannian manifolds $f: X \to Y$ that pulls back the latter's metric to the first. What happens on the Riemannian structure on the tangent bundle, then, is just $df$. The construction you're doing here is a vector bundle automorphism - it's not just $df$ where $f$ is the map $X \to X$ it lies over. If you could show that the $f: TX \to TX$ you're constructing was of the form $dg$ for some $g: X \to X$, that would show all Riemannian metrics were isometric. Of course, you can't.
– user98602
Commented Mar 28, 2015 at 6:38

Note that every positive definite map $\mu$ can be written uniquely as square of $\sqrt \mu$, i.e. $\mu=\sqrt \mu \sqrt \mu$.

If $\mu'$ is another positive definite map, then we want a positive definite map $M:E(\xi) \to E(\xi)$ which gives us an isometry. The isometry will be in fact the square root $\sqrt M$ of the following composition $M$:

Let $b_\mu :E(\xi) \to E(\xi)^*$ the map which sends a vector $v$ to the linear map $\mu(v,\cdot)$. Very basic linear algebra tells you this is an isomorphism, hence you can try to compose $b_\mu$ and $b_{\mu'}$ to get the map $M$. Intuitively speaking, you will go to the dual by applying $\mu'$ and going back by $\mu$, and the resulting isomorphism will 'measure the difference'.

Edit: To answer the commented questions: we need the square root of $M$ in the first place because of the bilinearity and our wish that $\mu'(\sqrt M v, \sqrt M w) = \mu (v,w)$. So you also see here how you want to define the square root obviously. To make this precise you could note that the function on $n$-matrices $M \mapsto M^2$ restricts to a function on the space of positive definite $n$-matrices to ifself. You want to show that it actually restricts to a homeomorphism, i.e. that we have an inverse map $$\{\text{positive definite n-matrices}\} \stackrel {((\cdot)^2)^{-1}}\to \{\text{positive definite n-matrices}\} \subset GL_n(\mathbb R)$$ which gives us the square root map.

Note that this is not a complete proof, since your question was only about how square roots come into play here. Note also, that most of the work is linear algebra, but to translate it to vector bundles you need to worry a little about continuity.

$$\sqrt{M: E(\xi) \stackrel {b_{\mu'}} \to E(\xi)^* \stackrel {b_\mu} \to E(\xi)}$$

• I'm still a little confused on how the square roots work - why doesn't $M$ just give you what you want? Also, is the square root with respect to multiplication of matrices (i.e. composition) or with respect to pointwise multiplication in the target? Commented Mar 27, 2015 at 23:50
• I edited my post to answer your questions. Commented Mar 28, 2015 at 0:05
• I apologize, but there's something (probably silly) I'm still not getting - I don't see how the relation $\sqrt{M} \circ \sqrt{M} v = Mv$ is the same as $\mu'(\sqrt{M} v, \sqrt{M} w) = \mu(v,w)$ $\mu'$ is linear, but it doesn't make sense to pull the matrices out of $\mu'$, does it? (since then everything is just one dimensional) Commented Mar 28, 2015 at 5:57