Let $f: M \to N$ be a smooth map between smooth manifolds.

Two vector fields $X$ in $M$ and $Y$ in $N$ are said to be $f$-related if for all $p \in M$, $(f_*)_p(X_p) = Y_{f(p)}$; equivalently, if for every smooth function $g: N \to \mathbb{R}$, one has $$(Yg) \circ f = X(g \circ f).$$

One immediate consequence of this definition is that if $X_i$ and $Y_i$ are $f$-related, for $i=1,2$, then so are their Lie brackets $[X_1,X_2]$ and $[Y_1,Y_2]$.

A number of basic results about Lie groups in a differential-geometric context follow from this observation; for example,

  • the left-invariant/right-invariant vector fields form a Lie subalgebra,
  • the differential of a smooth map between Lie groups is a Lie algebra homomorphism,
  • the Lie algebra of a Lie subgroup is a Lie subalgebra (really a special case of the above),
  • etc

Now, perhaps unwisely, while I was preparing a problem sheet for an undergraduate summer project, I added a problem on f-relatedness, one of whose parts was to show the following.

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$, acting on the left right on a smooth manifold $M$. If $X \in \mathfrak{g}$, let $X'$ denote the corresponding fundamental vector field on $M$. Show that $[X',Y'] = [X,Y]'$, where the left-hand side is the Lie bracket of vector fields on $M$ and the right-hand side is the vector field on $M$ corresponding to the Lie algebra bracket of $X,Y \in \mathfrak{g}$.

One can prove this directly, of course, but I (mistakenly?) thought there ought to be a slick proof using $f$-relatedness. After one of my students complained that he could not find a proof using $f$-relatedness, I realised (somewhat embarrassingly) that neither could I!

The naive thing does not work: one can exhibit $X'_m = (\alpha_m)_* X$, where $\alpha_m : G \to M$ is the map sending $g$ to $g\cdot m$. This does not work because the map used to relate $X$ and $X'$ depends on $m$.

Hence the following


Is the vector field $X'$ on $M$ $f$-related to the left-invariant vector field on $G$ corresponding to $X \in \mathfrak{g}$?

In other words, the question is to exhibit $f$. It does not seem that it's as simple as a map $G\to M$, but perhaps there is some trick using the action $\alpha: G \times M \to M$ in some way.

  • 1
    $\begingroup$ If $\psi: G \to Diff(M)$ is the action, then isn't $X' = \psi_* X$, where we identify $Lie(Diff(M))$ with vector fields on $M$? $\endgroup$ Commented Jul 13, 2010 at 0:26
  • $\begingroup$ Actually I think you're right, Eric! In fact, this is probably what I had in the back of my mind... Why don't you write your comment as an answer? If no other answer appears involving only finite-dimensional manifolds, I would accept it and in any case I'm happy to give it an upvote. $\endgroup$ Commented Jul 13, 2010 at 0:37

3 Answers 3


I hate to throw cold water on the party, but surprisingly, the formula that the OP was trying to prove ($[X',Y']=[X,Y]'$) is actually false when $G$ acts on the left. This formula is correct if $G$ acts on the right on $M$; but if $G$ acts on the left, then the correct formula is $[X',Y']=-[X,Y]'$.

Here's how to prove it. Suppose first that $G$ acts smoothly on $M$ on the right. Fix $p\in M$, and consider the orbit map $\alpha^{(p)}: G\to M$ defined by $\alpha^{(p)}(g) = p\cdot g$. Then I claim that for each $X\in \mathfrak g$, the fundamental vector field $X'$ is $\alpha^{(p)}$-related to $X$. To see this, note that the group law $p\cdot gg' = (p\cdot g)\cdot g'$ translates to $$\alpha^{(p)}\circ L_g (g') = \alpha^{(p\cdot g)}(g')$$ (where $L_g$ is left multiplication by $g$). Let $g\in G$ be arbitrary and set $q=p\cdot g=\alpha^{(p)}(g)$. Because $X$ is left-invariant, $$ X'_q = d\bigl(\alpha^{(q)}\bigr)_1(X_1) = d\bigl(\alpha^{(p)}\bigr)_g\circ d(L_g)_1(X_1) = d\bigl(\alpha^{(p)}\bigr)_g(X_g)$$ (where the first equality is essentially the definition of $X'$, and the second follows from the previous equation by taking differentials at the identity). This proves the claim.

Because brackets of $\alpha^{(p)}$-related vector fields are themselves $\alpha^{(p)}$-related, it follows that $[X',Y']_p = ([X,Y]')_p$, and since this is true for every $p$ it is true globally.

Now if $G$ acts on $M$ on the left, the above argument doesn't work; if $\tilde\alpha^{(p)}$ denotes the orbit map for the left action, then $X'$ is not $\tilde\alpha^{(p)}$-related to $X$. Instead, we can create a right action by setting $p\cdot g = g^{-1}\cdot p$. Letting $\alpha^{(p)}$ denote the orbit map for this new right action, we have $\tilde\alpha^{(p)}=\alpha^{(p)}\circ\iota$, where $\iota: G\to G$ is inversion; and the argument above shows that $X$ and $X'$ are $\alpha^{(p)}$-related. Since $d\alpha^{(p)}$ preserves brackets and $d\iota$ reverses them, it follows that $d\tilde\alpha^{(p)}$ reverses brackets.

The problem with Eric's argument has to do with the identification between $\operatorname{Lie}(\operatorname{Diff}(M))$ and $\mathfrak X(M)$. Essentially the same argument as above shows that the natural map $\operatorname{Lie}(\operatorname{Diff}(M)) \to \mathfrak X(M)$ is actually a Lie algebra anti-isomorphism (if we consider $\operatorname{Diff}(M)$ as an infinite-dimensional Lie group acting on $M$ on the left).

The problem with Victor Protsak's argument is that it doesn't actually show that $(X,0)$ (considered as a vector field on $G\times M$) is $\alpha$-related to $X'$ -- to show this, you'd have to prove that $(d\alpha)_{(g,m)}(X,0)=X'$ for every $g$ in the group, not just $g=1$.

I had to sort all this out recently because I'm adding a section on infinitesimal group actions in the second edition of my book Introduction to Smooth Manifolds. When I tried to prove the same formula the OP was trying to prove, I was surprised to find that it led to contradictions. Eventually I came up with the argument I sketched here, and then found at least one other book that confirms the result. I'm at home now and don't have access to my books, but if you'd like I'll find the reference and post it tomorrow.

  • $\begingroup$ Thank you for pointing out that $\alpha$-relatedness requires considering all elements $(g,m)$. I am less confident about your conclusion, because to me it's second nature to replace the Lie group action on a manifold with a homomorphism, and not antihomomorphism, of Lie algebras $\mathfrak{g}\to \operatorname{Vect}(M).$ (But I have not thought it through for many years; there are certainly some tricky points in identifying Lie algebras with vector fields because of the left-right issues. Perhaps, your definition of the fundamental vector field differs by the sign.) $\endgroup$ Commented Jul 13, 2010 at 6:37
  • $\begingroup$ I was going to say that in my definition of fundamental vector field, I always get a homomorphism, by construction. But then I define the value at $m$ of the fundamental vector field $X'$ as the velocity of the curve $e^{-tX}\cdot m$. But then why do I call it a left action?! So, my apologies. I do mean a right action. In fact, this is why on a Lie group, it's the right-invariant vector fields which generate left translations, and viceversa. $\endgroup$ Commented Jul 13, 2010 at 12:03

Since I could use the reputation... letting $\psi: G \to \operatorname{Diff}(M)$ be the action then $X' = \psi_* X$, where we identify $\operatorname{Lie}(\operatorname{Diff}(M))$ with vector fields on $M$. This is because they both correspond to the one-parameter subgroup $t\mapsto \psi(\exp tX)$ of $\operatorname{Diff}(M)$. Then $[X', Y'] = [X,Y]'$ is immediate.

  • $\begingroup$ Thanks. The more I think about it, I'm convinced that this was what prompted me to set that problem. It's funny how sometimes one knows something is true, but then has difficulty reproducing the argument. I guess it's age... :( $\endgroup$ Commented Jul 13, 2010 at 1:35
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    $\begingroup$ This is just me being pedantic about TeX. I made only superficial edits to your comment to include the command \operatorname, so that TeX knows that Diff should not be the awkwardly spaced $Diff = D \times i \times f \times f$, but rather $\operatorname{Diff}$, an operator like $\exp, \log, \sin$ and so on. +1 for the answer, BTW. $\endgroup$ Commented Jul 13, 2010 at 5:23

The fact you are trying to show follows by considering the differential of $\alpha: G\times M\to M$ at $\{1\}\times M.$ After the identification $T_{(1,m)}(G\times M)\simeq T_1 G\oplus T_m M,$


or in your language, vector fields $(X,0)$ and $X^\prime$ are $\alpha$-related. Then

$$ d\alpha([(X,0),(Y,0)])=[X,Y]'\ \text{ and }\ [d\alpha(X,0),d\alpha(Y,0)]=[X',Y'], \text{ so } [X,Y]'=[X',Y'].$$

  • $\begingroup$ Thanks-- seems like this is it. I wish I could accept both yours and Eric's answers. In the present context -- an undergraduate project -- I think that this answer is more self-contained, so I will accept it. Perhaps I should have thought longer about this question... the mind plays tricks. $\endgroup$ Commented Jul 13, 2010 at 2:07
  • $\begingroup$ You are welcome, José! There are some foundational issues with treating $Diff(M)$ as a Lie group, but heuristically, the two aproaches are equivalent due to the "internal hom" property $\operatorname{Hom}(G\times M,M)\simeq \operatorname{Hom}(G,\operatorname{Hom}(M,M)).$ $\endgroup$ Commented Jul 13, 2010 at 2:12
  • $\begingroup$ As Jack Lee pointed out in his answer, in order to conclude $\alpha$-relatedness one has to consider $(d\alpha)_{g,m}$ at all $g$, not just $g=1.$ I am going to sleep on it, though. $\endgroup$ Commented Jul 13, 2010 at 6:26

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