Prove that $\{A \in M_{2x2}: \text{rank}(A)=1\}$ is a 3-submainfold

Question

First of all, English is not my native language so I may use some wrong terms, if so, let me know.

I'm studying submanifolds, and I need to prove that $$N:=\{A \in M_{2x2}: \text{rank}(A)=1\}$$ is a submanifold of $M_{2x2}$ with dimension 3.

The only way I can think of is defining $f(A)=\det(A)$ so that if $N$ were $N=f^{-1}(\{0\})$ and $\nabla f(p)\neq 0 \enspace\forall p \in N$, then this would be proved.

However, that's obviously not the case, as $0\in f^{-1}(\{0\}), \enspace 0\notin N$ and $\nabla f(0)=0$.

I have tried to think any other function to solve this, for example $g(x, y, z, w) = (x*z - y*w, x^2+y^2+z^2+w^2)$ so that $N=g^{-1}(\{(0, x): x\neq 0\})$ but that proves nothing.

Is there any function I could use to solve this? Or what method should I use?

Thanks a lot!!

Answer 1

You can fix your method's problems with the zero matrix by noting that $M_{2 \times 2} \setminus \{0\}$ is also a four-dimensional manifold, since it is the four-dimensional manifold $M_{2 \times 2}$ with just a single point removed.

Then, take the function $f: M_{2 \times 2} \setminus \{0\} \rightarrow \mathbb{R}$ given by the determinant on this restricted domain. If we look at the preimage $f^{-1} (0)$ we see that:

• There isn't any problem with the derivatives. The only place where the derivative could be zero is at the zero matrix, which isn't in the domain. All points in the domain have non-zero derivative, so this preimage is a three-manifold.
• Since the preimage doesn't contain zero (which isn't in the domain) it only contains those matrices which aren't zero but don't have full rank. Since we're only working with 2-by-2 matrices, that means that the matrices in the preimage have rank 1. This makes it exactly the set you are looking for.

Answer 2

Here is a possibility to exhibit (explicitly) the structure of a manifold for the set $$ M = \left\{\ \begin{bmatrix} a & b \\ c & d\end{bmatrix}\ : \ a,b,c,d\in\Bbb R\ ,\ \operatorname{rank} \begin{bmatrix} a & b \\ c & d\end{bmatrix}=1\ \right\} \ . $$ I will also write $(a,b;c,d)$ instead of the matrix with entries $a,b,c,d$ to have a line fit and easy type. The matrices in $M$ have determinant zero, so this implies $ad-bc=0$. This condition, together with the condition that the rank does not vanish, i.e. not all variables are zero, characterizes an element of $M$. We can thus cover $M$ using four sets, $$ \begin{aligned} U_a &=\{\ A=(a,b;c,d) \in M\ :\ a\ne 0\ \}\ ,\\ U_b &=\{\ A=(a,b;c,d) \in M\ :\ b\ne 0\ \}\ ,\\ U_c &=\{\ A=(a,b;c,d) \in M\ :\ c\ne 0\ \}\ ,\\ U_d &=\{\ A=(a,b;c,d) \in M\ :\ d\ne 0\ \}\ , \end{aligned} $$ and in each case we can write a map to $\Bbb R^3$ which is homeo when restricted to the image. For instance, on $U_a$ we have $ad=bc$, so the component $d$ is explicitly determined dy the other three components, and we can use the chart given by the following map and its inverse: $$ \underbrace{\Bbb R^*\times\Bbb R\times \Bbb R}_{:=V_a}\to U_a\ ,\ (a,b,c)\to\begin{bmatrix}a&b\\c&\frac{bc}a\end{bmatrix}\ . $$ (For $U_a$ we "supress", omit the "$d$"-component, which is determined by division with the "$a$"-component, and do correspondingly for the other components.)

It is clear that $U_a,U_b,U_c,U_d$ are covering $M$ (since for any $A=(a,b;c,d)$ of $M$ at least one of its components does not vanish, so $A\in U_a$ if $a\ne 0$, $A\in U_b$ if $b\ne 0$, etc. .

The analytic and/or smooth and/or $C^k$ stucture from $V_?\subset\Bbb R^3$ is then transfered to $U_?\subset M$, $?\in\{a,b,c,d\}$, and we have to show that the compositions of charts (and inverse charts) like $V_{ab}\to V_{ba}$ are in the same class. (Analytic, smooth, $C^k$.)

Here, let us denote by $U_{ab}=U_{ba}=\{\ (a,b;c,d)\ :\ a,b\ne 0\ \}$ the intersection $U_a\cap U_b$, then further denote

• by $V_{ab}$ the preimage of $U_{ab}$ via the structure chart $V_a\to U_a$, and
• by $V_{ba}$ the preimage of $U_{ba}=U_{ab}$ via the structure chart $V_b\to U_b$.

We use appropiate notations and meanings for the other open domains in $\Bbb R^3$ and maps between them.

Using the "flip symmetries" of $M$, e.g. $(a,b;c,d)\to(d,b,c,a)$, and (the transposition) $(a,b;c,d)\to(a,c,b,d)$ and compositions of them, it is enough to study the maps:

• $V_{ad}\to V_{da}$, and
• $V_{ab}\to V_{ba}$.

The first map above is defined for $(a,b,c)\in V_{a}$ with $a,b,c\ne 0$, and it is $(a,b,c)\to (a,b;c,bc/a)\to (b,c,bc/a)$. (We have not specified explicitly the last map, up to permutation of its components it is but the above one.) This map is analytic.

The second map above is defined for $(a,b,c)\in V_a$ with $a,b\ne 0$ and it maps $(a,b,c)\to(a,b;c,bc/a)\to\dots$ and from the last element we use $b\ne 0$ to supress the $c$ component, so $\dots\to(a,b,bc/a)$, again an analytic map.