Suppose $M_1, \dots M_k$ are smooth manifolds, then each of the projection maps $\pi_i : M_1 \times \dots \times M_k \to M_i$ is a submersion

Question

Suppose $M_1, \dots M_k$ are smooth manifolds of dimensions $n_1, \dots, n_k$ respectively, then each of the projection maps $\pi_i : M_1 \times \dots \times M_k \to M_i$ is a smooth submersion

This was my attempted proof:

Proof: Let $M = M_1 \times \dots M_k$ and let $p = (p_1, \dots, p_k) \in M$ choose a smooth chart $(U, \phi)$ containing $p$. Then $U = U_1 \times \dots \times U_k$ and $\phi = \phi_1 \times \dots \times \phi_k$ where each $(U_j, \phi_j)$ is a smooth chart in $M_j$. Also note that $\phi = \phi_1 \times \dots \times \phi_k$ means that $\phi(p_1, \dots, p_k) = (\phi_1(p_1), \dots, \phi_k(p_k)) = \left(\phi_1 \times \dots \times \phi_k\right)(p_1, \dots, p_k)$.

Let's now compute the local coordinate representation of $\widehat{\pi_i}$, this is given by $\widehat{\pi_i} = \phi_i \circ \pi_i \circ \phi^{-1} : \phi[U] \to \phi_i[U_i]$. Observe that $\phi^{-1} = \phi_1^{-1} \times \dots \times \phi_k^{-1}$. Letting $(x^1, \dots, x^{n_i})$ denote the component functions of $\phi_i$ we have \begin{align*}\widehat{\pi_i}(x^1, \dots, x^{n_1 + \dots + n_k}) &= \phi_i\bigg(\pi_i\big(\phi_1^{-1}(x^1, \dots, x^{n_1 + \dots + n_k}), \dots, \phi_k^{-1}(x^1, \dots, x^{n_1 + \dots + n_k})\big)\bigg) \\ &=\phi_i\bigg( \phi_i^{-1}(x^1, \dots, x^{n_1 + \dots + n_k})\bigg) \\ &=(x^1, \dots, x^{n_1 + \dots + n_k}) \end{align*}

Then $d\widehat{\pi}_{\phi(p)}$ is represented by the Jacobian matrix of $\widehat{\pi_i}$ at $\phi(p)$ and computing the Jacobian matrix for $\widehat{\pi_i}$ we see that $$d\widehat{\pi}_{\phi(p)} = \begin{bmatrix} I & 0 \\ 0 & 0\\ \end{bmatrix}$$ with $n_i$ many $1$'s down the main diagonal. Thus since $d(\pi_i)_p = d(\widehat{\pi_i})_{\phi(p)}$ in the chart $(U, \phi)$ we have $\operatorname{rank}d(\pi_i)_p = \operatorname{rank}d(\widehat{\pi_i})_{\phi(p)} = n_i = \operatorname{dim}T_{p_i} M_i$ hence $d(\pi_i)_p$ is surjective and thus $\pi_i$ is a smooth submersion. $\square$

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Is my proof correct? If not where have I gone wrong?

Answer 1

It is almost correct. The idea is right, but I would have written it like this:

Let $\mathbf x_i \in \mathbf R^{n_i}$, so $\mathbf x = (\mathbf x_1,\dots,\mathbf x_k)$ is a point in $\mathbf R^{n_1}\times\dotsb\times\mathbf R^{n_k}$.

Then, \begin{align*} \widehat\pi_i(\mathbf x) = \mathbf x_i, \end{align*} or in matrix form as $$ \widehat\pi_i = \begin{bmatrix} 0 & \dotsb & I_i & \dotsb & 0 \end{bmatrix}, $$ where $I_i$ is the $n_i\times n_i$ identity matrix. Since $d\widehat \pi_i = \widehat \pi_i$, and $\widehat\pi_i$ clearly has rank $n_i$, $\widehat \pi_i$ is a submersion.

Note, in particular, what is wrong with your proof:

• You have written that the coordinate form of $\pi_i$ is actually the identity map $\mathbf R^{n_1}\times\dotsb\times\mathbf R^{n_k}\to\mathbf R^{n_1}\times\dotsb\times\mathbf R^{n_k}$ by just not being careful.

• The matrix for $\widehat \pi_i$ is not correct, and actually shows the wrong conclusion because the matrix you wrote down does not have the correct dimensions. You should have found that the coordinate representation is a map $\mathbf R^{n_1}\times\dotsb\times\mathbf R^{n_k}\to \mathbf R^{n_i}$, so the matrix should have dimensions $n_i\times(n_1+\dotsb+n_k)$.