# Representation of $\mathrm{Diff}(M)$ onto $C^\infty(M)$

I came up with the following question when studying symmetries in physics. Please note that some part of my question is not precise.

Let $$M$$ be a smooth manifold, and let $$\mathrm{Diff}(M)$$ be the diffeomorphism group of $$M$$. Consider $$\mathrm{Diff}(M)$$ as an (infinite-dimensional) Lie group. Then, we have a natural Lie group representation of $$\mathrm{Diff}(M)$$ on the space of smooth functions $$C^\infty(M)$$ by $$\rho: \mathrm{Diff}(M) \to \mathrm{GL}(C^\infty(M)),$$ where $$\rho(F)(f) = f \circ F^{-1}$$. In other words, a diffeomorphism $$F:M \to M$$ is regarded as a coordinate transform. Here, we have to consider the inverse $$F^{-1}$$ in the definition of $$\rho$$ to ensure that $$\rho$$ is a representation, i.e., $$\rho(F\circ G) = \rho(F) \circ \rho(G)$$.

Question: Lie group representation induces the corresponding Lie algebra representation. What is the Lie algebra representation corresponding to $$\rho$$?

Some thoughts: First, we should determine what is the Lie algebra of $$\mathrm{Diff}(M)$$. I suspect that it is $$\mathfrak X(M)$$, the space of vector fields on $$M$$, since vector field encodes information of the infinitesimal transformation. Then, for $$X\in \mathfrak X(M)$$, considering (formally) $$e^{tX} \in \mathrm{Diff}(M)$$, where $$t\in\mathbb R$$ is infinitesimal, we write $$(e^{tX} \cdot f)(x) := \rho(e^{tX})(f)(x) = f(e^{-tX}x),$$ where we used the simplified notation $$e^{tX} \cdot f$$ to denote the representation $$\rho$$. By Taylor expanding both sides and considering the terms up to $$O(t)$$, we have $$f(x) + t (X\cdot f)(x) = f(x^a - tX^a) = f(x) - t X^a \partial_a f(x).$$ Hence, I expect $$X\cdot f = -X^a \partial_a f.$$ However, the minus sign here is somewhat strange. Is this a Lie algebra representation? (If there were no minus sign, that the above is indeed the Lie algebra representation.) I cannot figure out where is wrong.

• Yes the natural Lie algebra of $\mathrm{Diff}(M)$ is the set of vector fields on $M$ but we should be wary because these are infinite dimensional and all of our finite dimensional intuition may not go through. In this particular case we are okay however. The set of vector fields acts on $C^\infty(M)$ as derivations (arguably by definition) and this is our Lie algebra representation. Oct 14, 2022 at 11:08
• @Callum Thanks for your comment, but I cannot see how $X\cdot f = -X^a \partial_a f$ is a Lie algebra representation. One may write the RHS $-X^a \partial_a f = \mathcal L_X f$ using the Lie derivative, then we have $X\cdot f = -\mathcal L_X f$. However, since $\mathcal L_{[X,Y]} = [\mathcal L_X, \mathcal L_Y]$, the additional minus sign prevsents us from being the Lie algebra representation. Oct 14, 2022 at 13:44

## 2 Answers

The corresponding Lie algebra representation "ought to be" the action of vector fields by the Lie derivative. The sign is supposed to be there. The issue, as far as I can tell, is that there is a sign convention in the definition of the Lie bracket on vector fields and to make it consistent with the conventions given here that sign should be reversed; that is, the definition of the Lie bracket on vector fields should be the negative of the commutator bracket.

Basically I think this comes down to whether you think a vector field $$X$$ should exponentiate to $$f(x) \mapsto f(e^{tX} x)$$ (the convention that defines the usual Lie derivative and the usual Lie bracket on vector fields) or $$f(x) \mapsto f(e^{-tX} x)$$ (the convention you are adopting here).

A few comments to expand on Qiaochu Yuan's answer: it is useful to distinguish between right group actions and left group actions. In your case, you are looking at a left action of the group of diffeomorphisms on the space of functions.

Consider a Lie group $$G$$, and let $$\mathfrak{g}:=\operatorname{Lie}(G)$$ be its Lie algebra; recall that this $$\mathfrak{g}=T_eG$$ is identified with the space of left-invariant vector fields on $$G$$.

For a left-action of $$G$$ on a manifold $$M$$, let's say $$x\to\lambda_g(x)$$, one can define an infinitesimal action of $$\mathfrak{g}$$ on $$M$$ as follows. To each $$a\in\mathfrak{g}=T_eG$$ we associate the vector field $$\hat{a}_x:=\partial_{t=0}\lambda_{\operatorname{exp(t\,a)}}(x).$$ This is a Lie algebra anti-homomorphism between $$\mathfrak{g}$$ and the set $$\mathcal{X}(M)$$ of vector fields on $$M$$: to get a Lie algebra homomorphism instead, you should define $$\hat{a}$$ as $$\hat{a}_x:=\partial_{t=0}\lambda_{\operatorname{exp(-t\,a)}}(x),$$ see for example Proposition $$3.8$$ in Appendix 5 of this book.

From this point of view then it is not surprising that you have a minus sign in your infinitesimal action. This can be corrected, if you need to do so, by either considering a right-action of $$G$$, or by changing the definition of Lie bracket by a minus sign (which is roughly the same as considering $$\mathfrak{g}$$ as the set of right-invariant vector fields).

• Oh, I did not know that the map $a \mapsto \hat a$ is a Lie algebra "anti"-homormorphism. After knowing this, then now I see why I had a minus sign. Thanks for your answer! Oct 15, 2022 at 1:58
• I read the cited book, and also searched "fundamental vector field" on Wikipedia. (en.wikipedia.org/wiki/Fundamental_vector_field) I found out that the definition on the Wikipedia has the different sign compared to the book. More precisely, the Wikipedia says $$X_p^\# = \left. \frac{d}{dt} \right|_{t=0} A(\exp(+tX), p)),$$ while the book (and your answer) has the essentially same expression with the minus sign $\exp(-tX)$. Does the Wikipedia have sign mistake? Oct 15, 2022 at 10:52
• @eigenvalue you are correct, it seems that Wikipedia is slightly confused about these matters. In the page you linked they indeed use $X^\sharp_p=\partial_{t=0}\mathrm{exp}(tX).p$, but in the page relative to group actions (en.wikipedia.org/wiki/Lie_group_action) they follow the same convention as in my answer, $X^\sharp_p=\partial_{t=0}\mathrm{exp}(-tX).p$. Given this discrepancy, I would refer to the book, which gives a proof of its claim.It could be that the author of the article you linked had right-actions in mind? Oct 15, 2022 at 16:16
• Thanks for your comments! Oct 16, 2022 at 23:57