Consider the subset of all 2x2 matrices where the determinant is 0 and trace of A transpose times A is 4 - is the set a manifold and if so what dimension manifold is it?

Answer 1

Let M be the subset of 2x2 matrices A such that det(A)=0 and tr(A'A)=4 (I shall use ' to denote transpose).

Recall that one of the three definitions of a k-dimensional manifold is:

M c R^n is a k-dimensional manifold if for any p in M there is a neighborhood W c R^n of p and a smooth function F:W-->R^n-k so that F^-1(0) = M intersection W and rank(DF(x))=n-k for every x in M intersection W.

In short, what we need to do is find a function F so that the inverse image of the zero vector under F gives M and the rank of the derivative of F is equal to the dimension of the codomain of F.

Let an arbitrary 2x2 matrix be written as:

[a c]

[b d]

Then the two constraints that define M are

1) ad-bc=0

2) a^2+b^2+c^2+d^2=norm(a,b,c,d)^2=4

Define F:R^4-->R^2 by F(a, b, c, d)=(ad-bc, norm(a,b,c,d)^2-4). Then clearly F^-1(0,0)=M. Furthermore, F is smooth on M because the multiplication and addition of smooth functions (a, b, c, d) is also smooth. (Note that we have taken the neighborhood W to be some superset of M. This guaranteed to exist because if we interpret the set of 2x2 matrices as R^4, every point in M has norm 2, so any ball centered at the origin with length greater than 2 will contain M).

All that remains to be done is to check that rank(DF(x))=2 for every x in M. Observe that [DF]= [d -c -b a]

[2a 2b 2c 2d]

We shall now argue by contradiction. Suppose rank(DF) did not equal 2 for every x in M. Then we know that the two rows are linearly dependent i.e.

h[d -c -b a] + k[2a 2b 2c 2d] = 0 and h and k are not both 0.

Suppose h is 0. Then we have 2ka = 2kb = 2kc = 2kd = 0, and since k cannot also be 0, this implies that a=b=c=d=0, therefore norm(a,b,c,d)^2=0. But, (a,b,c,d) must be in M, so this is a contradiction. Hence h cannot be 0.

Now suppose h is nonzero. Then we can divide it out and there exists a, b, c, d and a constant k so that

[d -c -b a] + k[2a 2b 2c 2d] = 0 i.e. we have:

d + 2ka = -c + 2kb = -b + 2kc = a + 2kd = 0. From this we can substitute to obtain:

a(1 - 4k^2) = b(4k^2 - 1) = c(4k^2 - 1) = d(1 - 4k^2) = 0 and hence

a^2(1 - 4k^2)^2 = b^2(4k^2 - 1)^2 = c^2(4k^2 - 1)^2 = d^2(1 - 4k^2)^2 = 0.

Note that (1 - 4k^2)^2 = (4k^2 - 1)^2. Now, adding the four above expressions together we get:

(a^2 + b^2 + c^2 + d^2)(1 - 4k^2)^2 = norm(a,b,c,d)^2(1 - 4k^2)^2 = 0. But, since we require (a,b,c,d) to be in M, this reduces to 4(1 - 4k^2)^2 = 0. This implies that

1 - 4k^2 = 0, and hence k = +/-(1/2).

Now, if k=+1/2, then we have a + d = b - c = 0, therefore d = -a and b = c. Since we have det(A) = 0 as one of our constraints on M, this implies that ad - bc =

-(a^2) - (b^2) = -(a^2 + b^2) = 0, which implies a = b = 0, by the property of norms. But, if a = b = 0, then (a,b,c,d) = (0,0,0,0) and hence norm(a,b,c,d)^2 = 0, which is a contradiction.

We can argue analogously for the case where k = -1/2. Hence, assuming that rank(DF) is not 2 for some (a,b,c,d) in M leads to a contradiction, so we conclude that rank(DF)=2 for all x in M.

Finally, from this result, we conclude that since F:R^4-->R^2=R^(4-2), M must be a 2-dimensional manifold.