How to show that is a submanifold or How to derivate the determinant function?

Question

I am trying to show that the space of $2\times 2$ matrix with rank equals $1$ is a submanifold of $\mathbb{R}^4 - \{0\}$ whoose the dimension equals $3$. To do this, I have defined $\det : \mathbb{R}^4 \to \mathbb{R}$, then the result will follows from the fact that this submanifold is precisely $\det ^{-1}(0)$. But for this, I have to show that on this set, the derivative of $\det$ has constant rank.

How can I do this?

Can you compute the differential of the determinant map? Write it out element, by element and compute the appropriate derivatives. — Michael Burr, Apr 20 '16 at 17:40
Your space is not precisely $\det^{-1}(0)$, because the zero matrix is there and it has rank zero. — Matthew Leingang, Apr 20 '16 at 17:41
@MatthewLeingang, sho it is enough to consider the entire $\mathbb{R}^4?$ — L.F. Cavenaghi, Apr 20 '16 at 17:42
@MichaelBurr, the problem is precisely the fact that I have no idea how to derivate the determinant... — L.F. Cavenaghi, Apr 20 '16 at 17:42
As long as $\det$ has constant rank on nonzero matrices, everything should be OK because the complement of the zero matrix is an open set in $M_{2\times 2}$. — Michael Burr, Apr 20 '16 at 17:43
@MichaelBurr, but if I consider the entire $\mathbb{R}^4$ it is also ok, right? So the problem remais in the fact that I cannot show that the derivative has constant rank... — L.F. Cavenaghi, Apr 20 '16 at 17:44

Michael Burr · Accepted Answer · 2016-04-20T17:54:31.110

2

Hint: Consider the matrix: $$ M=\begin{bmatrix} a&b\\c&d \end{bmatrix}. $$

Then the $\det$ map is $(a,b,c,d)\mapsto (ad-bc)$. Therefore, $\det_\ast=\begin{bmatrix} d,-c,-b,a\end{bmatrix}$, which is a surjective map from $T_MM_{2,2}\rightarrow T_{\det(M)}\mathbb{R}$, except when $a=b=c=d=0$.

Therefore, in each neighborhood of a matrix $M$ with rank $1$, the submersion theorem applies. Since an open subset of a manifold is a manifold, we know that $M_{2,2}\setminus\{0\}$ is a manifold, and we can work on that (so that the zero matrix is not a problem).

edited Apr 20 '16 at 17:54

answered Apr 20 '16 at 17:44

Michael Burr

32,867

now I got it. Thanks a lot... – L.F. Cavenaghi Apr 20 '16 at 17:51

How to show that is a submanifold or How to derivate the determinant function?

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