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I'm trying to show that the maps defined in https://math.stackexchange.com/a/413846/500094 are mutual inverses. I have problem with one direction:

$(g\circ f)(v)=g(d(\pi_X)_{(p,q)}v,d(\pi_Y)_{(p,q)}v)=d(\iota_X)_p(d(\pi_X)_{(p,q)}v)+d(\iota_Y)_q(d(\pi_X)_{(p,q)}v)=\\d(\iota_X\circ\pi_X)_{(p,q)}v+d(\iota_Y\circ\pi_Y)_{(p,q)}v=v+v=2v\ne v$

At which point am I mistaken?

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  • $\begingroup$ It's not true that $d(\iota_X\circ\pi_X)_{(p,q)}$ is the identity. (Though, it is true that $d(\pi_X \circ \iota_X)$ is!). It can't be the identity, since, by the chain rule, $d(\iota_X\circ \pi_X) = d\iota_X \circ d \pi_X$, so the map from $T_{p,q} X\times Y$ to itself factors through the smaller dimensional $T_{p}X$. More concretely, if $v$ is tangent to $\{p\}\times Y$, then $d(\iota_X \circ \pi_X)) v = 0$ since already $d\pi_X(v) = 0$. $\endgroup$
    – Jason DeVito
    Jan 22, 2018 at 20:32
  • $\begingroup$ $d(\iota_X\circ \pi_X)_{(p,q)}$ is projection onto the subspace $T_{(p,q)}(X\times \{q\})$. This might help. $\endgroup$
    – AnonymousCoward
    Jan 22, 2018 at 20:35
  • $\begingroup$ Geometrically, at any given point $(p,q)$ you can write any tangent vector uniquely as a sum of vectors orthogonal to $p\times Y$ and orthogonal to $X\times q$. The inverse simply takes two tangent vectors on $X$ at $p$ and on $Y$ at $q$ and sends them to the sum of these two orthogonal component vectors. The first observation says this is an isomorphism. $\endgroup$
    – Pedro ♦
    Jan 22, 2018 at 20:47

1 Answer 1

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In general, if $W$ is a f.d. vector space and $P_V\colon W \to W$ is a projection map with image $V\subset W$ and kernel $U$, then $W \cong U \oplus V$ and the map $P_U+P_V = \text{id}_W$ (where $P_U\colon W \to W$ is the projection map with image $U$).

In this case, take $L = d(\iota_X\circ \pi_X)_{(p,q)}$.

  1. Show $L$ is a projection map ($L\circ L = L$).

  2. What is the image of $L$?

  3. What is the kernel of $L$?

  4. What map appearing in your question above is a formula for projection onto the kernel of $L$?

If you can answer these questions, you will have a nice answer to the problem.

Update: here are some extra details:

  1. We can use properties of pushforward and the fact that $\pi_X\circ \iota_X = \text{id}_X$ to show that $$ L\circ L = d(\iota_X\circ \pi_X)_{(p,q)}\circ d(\iota_X\circ \pi_X)_{(p,q)} = d(\iota_X\circ (\pi_X\circ \iota_X)\circ \pi_X)_{(p,q)} = d(\iota_X\circ \pi_X)_{(p,q)} = L$$ so $L$ is a projection map.

  2. The image of $d(\iota_X\circ \pi_X)_{(p,q)}$ is the subspace $V = T_{(p,q)}X\times \{q\}$, the tangent space at $(p,q)$ of the submanifold $X\times \{q\}$. In short, the image of this map is the set of tangent vectors at $(p,q)$ to paths of the form: $$ \iota_X\circ \pi_x\circ \gamma: \mathbb{R} \to X\times Y$$ where $\gamma$ is a path in $X\times Y$ with $\gamma(0) = (p,q)$. Such a $\gamma$ can be written as $\gamma(t) = (\gamma_1(t),\gamma_2(t))$ for $\gamma_1\colon \mathbb{R}\to X$ and $\gamma_2\colon \mathbb{R}\to Y$. You should verify that $$\iota_X\circ \pi_x\circ \gamma(t) = (\gamma_1(t),q)$$ Thus the image of $d(\iota_X\circ \pi_X)_{(p,q)}$ is the set of vectors in $T_{(p,q)}X\times Y$ that can be written in the form $$d(\iota_X\circ \pi_X)_{(p,q)}\gamma'(0) = \frac{d}{dt}|_{t=0}(\gamma_1(t),q) = (\gamma_1'(0),0)$$ which is the desired tangent space.

  3. The kernel of $d(\iota_X\circ \pi_X)_{(p,q)}$ is the subspace $U = T_{(p,q)}\{p\}\times Y$. I leave the justification to you.

  4. So which map above gives projection onto $U$? It's $d(\iota_Y\circ \pi_Y)_{(p,q)}$. The proof is similar to items 1. and 2. above. In particular, this map is NOT the identity map.
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  • $\begingroup$ I've spoiled the answer to the second bullet point in the comments above, but you should still try to prove the first bullet point. $\endgroup$
    – AnonymousCoward
    Jan 22, 2018 at 20:52
  • $\begingroup$ Thank you. For now, having verified that $d(\iota_X \circ \pi_X)_{(p,q)}$ is not the identity, tried to fix my proof directly, and arrived at another contradiction. The differential $d(\iota_X \circ \pi_X)_{(p,q)}$ is the map that sends $v=(v_1,v_2)$ to $(v_1,0)$, right? At the same time the second differential sends $v$ to $v$. So their sum sends $v$ to $(2v_1,v_2)$, which should not be the case... $\endgroup$
    – user557
    Jan 22, 2018 at 23:27
  • $\begingroup$ Let me expand the answer a bit for you. $\endgroup$
    – AnonymousCoward
    Jan 23, 2018 at 7:35
  • $\begingroup$ Isn't there a problem with the equation $d(\iota_X\circ \pi_X)_{(p,q)}\gamma'(0) = \frac{d}{dt}|_{t=0}(\gamma_1(t),q) = (\gamma_1'(0),0)$? You seem to be using the same symbol for different objects and/or making some identification. Namely, the $\gamma'(0)$ on the LHS is a tangent vector in $T_{(p,q)}(X\times Y$), whereas the RHS is not a tangent vector. The derivative does not even make sense unless you are working in some coordinate chart, but even in this case, it is not clear to me what identification(s) you are using. (cont. below)... $\endgroup$
    – Matematleta
    Aug 28, 2019 at 5:41
  • $\begingroup$ .... The most you can say is that for suitable $f:X\times Y\to \mathbb R,\ d(\iota_X\circ \pi_X)_{(p,q)}\gamma'(0)(f) = \frac{d}{dt}|_{t=0}f(\gamma_1(t),q).$ I am not claiming that the result is false, only that there is some explanation that I am missing here. $\endgroup$
    – Matematleta
    Aug 28, 2019 at 5:41

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