1
$\begingroup$

$\DeclareMathOperator{Diff}{Diff}$

Fix $\varphi_0 \in \Diff(M)$.

Claim: The "tangent space" $T_{\varphi _0}\Diff(M) = \{ X \circ \varphi_0 \mid X \in \Gamma(TM) \}$.

Justification: Pick a smooth family of diffeomorphisms $\varphi: (-\varepsilon, \varepsilon) \to \Diff(M): t \mapsto \varphi(t) = \varphi_t$. Then for each $p \in M$, there is a smooth path $\gamma_p: (-\varepsilon, \varepsilon) \to M: t \mapsto \varphi_t(p)$, so $$(\varphi'(0))(p) = \left.{d \over dt}\right|_{t=0} \varphi_t(p) = \gamma_p'(0) \in T_{\gamma_p(0)}M = T_{\varphi_0(p)}M.$$ Thus, a "tangent vector" at $\varphi_0$ is not a vector field (unless $\varphi_0 = \operatorname{id}$); it's a function $M \to TM$ sending $p$ to something in $T_{\varphi_0(p)}M$. Fortunately, such a function uniquely determines a vector field because it factors as $$M \xrightarrow{\varphi_0} M \to TM:p \mapsto \varphi_0(p) \mapsto \text{something} \in T_{\varphi_0(p)}M,$$ where the second map is an honest vector field. The claim is now justified.

(In the context of integrating time-dependent vector fields, all this is usually summarized by the equation $${d \over dt} \varphi_t = X_t \circ \varphi_t.$$

I'm aware that there's a rigorous definition of a smooth structure for the infinite-dimensional $\Diff(M)$, but I know nothing about it. I'm considering a one-parameter family of diffeomorphisms to be smooth iff the induced map $M \times (-\varepsilon, \varepsilon) \to M$ is smooth (I've been told this doesn't really coincide with the smooth structure on $\Diff(M)$, but I'm not sure how). Given this definition of a smooth path in $\Diff(M)$, we can certainly ask for its velocity. This crude analogy is all I have in mind when I say "tangent vector.")

One consequence of the claim is that there's a canonical identification of $T_{\varphi_0}\Diff(M)$ with $T_{\operatorname{id}}\Diff(M) = \Gamma(TM)$ and hence a canonical parallelization. For finite-dimensional Lie groups, we have two canonical parallelizations given by left and right translation, but neither is "better" than the other. My question is: What is going on with $\Diff(M)$? Morally speaking, why does it have a canonical parallelization? Does it coincide with the one induced by left or right translation or neither?

I suspect that this parallelization of $\Diff(M)$ is neither the one induced by left translation, nor that induced by right translation. In fact, I suspect that the moral reason for this parallelization is that $\Diff(M)$ acts on $M$; more generally, I think an action of a Lie group $G$ on an arbitrary manifold induces a parallelization of $G$ by the same principle above. However, I can't prove any of these claims, nor can I find any references. I'm sure they're abundant and I'm just not looking in the right places. Any suggestions would be appreciated.

$\endgroup$
1
$\begingroup$

I believe Kriegl & Michor discuss the tangent bundle of a diffeomorphism group in their text The Convenient Setting for Global Analysis.

Anyway, here's a proof using synthetic differential geometry (SDG) that the tangent space to the diffeomorphism group of a smooth space $M$ is the space of vector fields $\mathfrak{X}(M)$. There's a version of this argument in Anders Kock's text.

In SDG, we work in a smooth topos. For concreteness, let's take the Dubuc topos, into which the category of smooth manifolds embeds fully faithfully.

In our smooth topos, the real line $R$ is augmented with infinitesimal elements. In particular, the subspace $D := \{ d \in R \mid d^2 = 0 \}$ is treated as the "walking tangent vector", and we define the tangent bundle of a space $M$ as the mapping space $M^D$ of "infinitesimal curves" in $M$, with the projection given by evaluating at $0$. The tangent map $f_*: M^D \to N^D$ of any map $f: M \to N$ is just given by postcomposition with $f$.

Now, a vector field $X: M \to M^D$ is a section of the projection $\mathrm{eval}_0: M^D \to M$, and by currying the vector field is equivalent to a map $X: D \to M^M$ such that $X(0)=1_M$.

But $M^M$ is just the space of maps $M \to M$, and is a smooth monoid under composition. So a vector field is precisely a tangent vector to the mapping space $M^M$ at the identity: it's an infinitesimal transformation!

But what about if we want to consider the group of invertible maps $M \to M$? Well, we can check that in fact the tangent space is the same. Given a vector field $X: D \to M^M$, we can show that $X(-d)=X(d)^{-1}$ provided $M$ is infinitesimally linear (this is proved in Kock's book). So in fact the space of invertible maps $\mathrm{Diff}(M)$ shares the same tangent space as $M^M$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.