# The Unapologetic Mathematician

## The Tangent Space of a Product

Let $M^m$ and $N^n$ be smooth manifolds, with $M\times N$ the $m+n$-dimensional product manifold. Given points $p\in M$ and $q\in N$ we want to investigate the tangent space $\mathcal{T}_{(p,q)}M\times N$ of this product at the point $(p,q)$.

For some notation, remember that we have the projections $\pi_1:M\times N\to M$ and $\pi_2:M\times N\to N$. Also, if we have a point $q\in N$ we get a smooth inclusion mapping $i_q:M\to M\times N$ defined by $i_q(p)=(p,q)$. Similarly, given a point $p\in M$ we get an inclusion map $j_p:N\to M\times N$ defined by $j_p(q)=(p,q)$. These maps satisfy the relations

\displaystyle\begin{aligned}\pi_1\circ i_q&=1_M\\\pi_2\circ j_p&=1_N\\\pi_1\circ j_p&=p\\\pi_2\circ i_q&=q\end{aligned}

where the last two are the constant maps with the given values. We can thus use the chain rule to calculate the derivatives of these relations

\displaystyle\begin{aligned}\pi_{1*(p,q)}\circ i_{q*p}&=1_{\mathcal{T}_pM}\\\pi_{2*(p,q)}\circ j_{p*q}&=1_{\mathcal{T}_qN}\\\pi_{1*(p,q)}\circ j_{p*q}&=0\\\pi_{2*(p,q)}\circ i_{q*p}&=0\end{aligned}

These are four of the five relations we need to show that $\mathcal{T}_{(p,q)}M\times N$ decomposes as the direct sum of $\mathcal{T}_pM$ and $\mathcal{T}_qN$. The remaining one states

$\displaystyle 1_{\mathcal{T}_{(p,q)}M\times N}=i_{q*p}\circ\pi_{1*(p,q)}+j_{p*q}\circ\pi_{1*(p,q)}=L\circ(\pi_{1*(p,q)},\pi_{2*(p,q)})$

where $L(u,v)=i_{q*p}(u)+j_{p*q}(v)$ is a linear map from $\mathcal{T}_pM\times\mathcal{T}_qN$ to $\mathcal{T}_{(p,q)}M\times N$. The real content of the first four relations is effectively that

$\displaystyle (\pi_{1*(p,q)},\pi_{2*(p,q)})\circ L=1_{\mathcal{T}_pM\times\mathcal{T}_qN}$

That is, we know that $L$ is a right-inverse of $(\pi_{1*(p,q)},\pi_{2*(p,q)})$, and we want to know if it’s a left-inverse as well. But this follows since both vector spaces $\mathcal{T}_{(p,q)}M\times N$ and $\mathcal{T}_pM\times\mathcal{T}_qN$ have dimension $m+n$. Thus the tangent space of the product decomposes canonically as the direct sum of the tangent spaces of the factors. In terms of our geometric intuition, there are directions we can go “along $M$“, and directions “along $N$“, and any other direction we can go in $M\times N$ is a linear combination of one of each.

Note how this dovetails with our discussion of submanifolds. The projection $\pi_1:M\times N\to M$ is a smooth map, and every point $p\in M$ is a regular value. Its preimage $\pi_1^{-1}(p)$ is a submanifold diffeomorphic to $N$. The embedding realizing this diffeomorphism is $j_p$. The tangent space at a point $(p,q)$ on the submanifold $j_p(N)$ is mapped by $\pi_{1*(p,q)}$ to $\mathcal{T}_pM$, and the kernel of this map is exactly the image of the inclusion $j_{p*q}$. The same statements hold with $M$ and $N$ swapped appropriately, which is what gives us a canonical decomposition in this case.

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April 27, 2011 - Posted by | Differential Topology, Topology

## 3 Comments »

1. Wow. Just stumbled on this awesome blog. I’ve set myself a goal of reading all of your back entries so I can “catch up” to the current exposition. Thanks for the (clearly significant) effort you’ve put into this; it’s a great resource!

Comment by Chris | April 28, 2011 | Reply

2. […] obvious example of submersion is a projection from a product manifold. As we’ve seen, the determinant of this projection is always a surjection. In fact, it’s a projection […]

Pingback by Submersions « The Unapologetic Mathematician | May 2, 2011 | Reply

3. Agree with Chris. This is a great resource and would like to echo the thanks, John. Although I’m not at a sufficient level to grasp all the technical details, this blath provides a great “benchmark” to enable insight into this portion of the “world”. Much appreciated.

Comment by Jubayer | May 11, 2011 | Reply