Let $\pi : M \rightarrow M/G$ be the canonical projection, where $M$ is a manifold and $M/G$ is a quotient manifold.

Now, what can we say about $d \pi (p) : T_pM \rightarrow T_p(M/G)$? From my intuition I would say that elements in $T_p(M/G)$ consists of elements in equivalence classes defined by $T_p(G_xp)$. In case that this is true. Does anybody know how to show that $T_p(M/G)$ consists actually of these kind of elements?

  • 1
    $\begingroup$ The kernel of $d\pi$ is precisely the tangent space to the $G$-orbit of $p$ at $p$. $\endgroup$ Jul 28 '15 at 19:47

First, we need to be a little more careful about notation. For each $p\in M$, the map $d\pi_p$ is a surjective linear map from $T_pM$ to $T_{\pi(p)}(M/G)$ (not $T_p(M/G)$, which doesn't quite make sense).

The tangent space $T_{\pi(p)}(M/G)$ does not "consist actually of these kinds of elements." Depending on your preferred definition, it consists either of equivalence classes of smooth curves in $M/G$, or of derivations of the space of (germs of) smooth real-valued functions on $M/G$.

However, what's interesting about this case is that there is a canonical isomorphism between the quotient vector space $T_pM/T_p(G\centerdot p)$ and $T_{\pi(p)}(M/G)$, where $G\centerdot p$ is the orbit of $p$ under $G$. (I'm not sure what you meant by $G_xp$.) The isomorphism is induced by the surjective linear map $d\pi_p\colon T_pM \to T_{\pi(p)}(M/G)$, whose kernel (as noted by @Travis) is exactly $T_p(G\centerdot p)$. By standard linear algebra, $d\pi_p$ descends to an isomorphism from $T_pM/T_p(G\centerdot p)$ to $T_{\pi(p)}(M/G)$. Because this isomorphism is defined independently of any choices, we can canonically identify each element of $T_{\pi(p)}(M/G)$ with an element of the quotient space $T_pM/T_p(G\centerdot p)$. This is what people mean when they say that the tangent space to $M/G$ at $\pi(p)$ "is" $T_pM/T_p(G\centerdot p)$.

  • $\begingroup$ Could you suggest me where I can find a proof of $\text{ker}\{{d \pi_p}\} = T_p{(G.p)}$ ? $\endgroup$
    – jon jones
    Jun 12 '16 at 0:09
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    $\begingroup$ @jonjones: I don't know where that exact statement is proved, but all of the ingredients can be found in my Introduction to Smooth Manifolds, at least in the case that the action of $G$ is free and proper (which are the usual hypotheses for concluding that $M/G$ is a manifold). First, Proposition 21.7 shows that $G\centerdot p$ is an embedded submanifold of $M$. Next, $\pi$ is a smooth submersion by Theorem 21.10, and it follows from the definition of $\pi$ that $G\centerdot p$ is one of its level sets. Finally, Proposition 5.38 shows that $\operatorname{Ker} d\pi_p = T_p(G\centerdot p)$. $\endgroup$
    – Jack Lee
    Jun 13 '16 at 21:10

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