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Background

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

Question

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

# Lie group actions and f-relatedness

Background

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 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

Question

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.