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I worked out the answer for the 2 by 2 case as well.

First, when dealing with 2 by 2 matrices in general, a convenient variable change is:

a->(w+x)/\sqrt{2},d->(w-x)/\sqrt{2},c->(y-z)/\sqrt{2},b->(y+z)/\sqrt{2}.


Then a^2+b^2+c^2+d^2 = w^2+x^2+y^2+z^2. And the determinant (ad-bc) = (1/2)*(x^2+y^2-w^2-z^2).

(Aside: this set of coordinates lets you see for instance that the set of rank 1 matrices in the space of 2D matrices realized as R^4 is a cone over the Clifford torus, since x^2+y^2 = w^2+z^2 on a sphere x^2+y^2+w^2+z^2=r^2 implies x^2+y^2 = r^2/2 and w^2+z^2 = r^2/2, which are scaled equations for a flat torus)

Let r1^2 = x^2+y^2, r2^2 = w^2+z^2. (These are radial coordinates of two cylindrical coordinate systems filling out 4-space). Then the norm squared is:

(1/2)*(r1^2+r2^2 + \sqrt{ (r1^2+r2^2)^2 - (r1^2-r2^2)^2 })


When this is less than one, this corresponds to the region plotted below:

Note that each point in the r1,r2 picture corresponds to a different "torus", x^2+y^2=r1^2, w^2+z^2=r2^2.

We can now integrate over the shaded in region, \int_{region} dw dx dy dz.

This 4-D integral can be reduced to 2D using r1 and r2, since dx dy = 2π r1 dr1, dw dz = 2π r2 dr2:

(4π^2) \int_{region} dr1 dr2 r1 r2


Now, note that we can rewrite r2 in terms of r1. In particular, after some manipulation of our norm, the shaded in region is defined by r2^2 ≤ 2-2\sqrt{2}r1+r1^2=(\sqrt{2}-r1)^2. Hence r2≤ \sqrt{2}-r1, and we can evaluate the r2 integral:

(4π^2) \int_{r1=0}^\sqrt{2} dr1 r1 \int_{r2=0}^{\sqrt{2}-r1} r2 dr2
= (4π^2) \int_{r1=0}^\sqrt{2} dr1 r1 (\sqrt{2}-r1})^2/2
= (4π^2) (1/6)


This yields 2π^2/3, as Armin found.

3 minor edits

I worked out the answer for the 2 by 2 case as well.

First, when dealing with 2 by 2 matrices in general, a convenient variable change is:

a->(w+x)/\sqrt{2},d->(w-x)/\sqrt{2},c->(y-z)/\sqrt{2},b->(y+z)/\sqrt{2}.


Then a^2+b^2+c^2+d^2 = w^2+x^2+y^2+z^2. And the determinant (ad-bc) = (1/2)*(x^2+y^2-w^2-z^2).

(Aside: this set of coordinates lets you see for instance that the set of rank 1 matrices in the space of 2D matrices realized as R^4 is a cone over the Clifford torus, since x^2+y^2 = w^2+z^2 on a sphere x^2+y^2+w^2+z^2=R^2 x^2+y^2+w^2+z^2=r^2 implies x^2+y^2 = R^2/2 r^2/2 and w^2+z^2 = R^2/2r^2/2, which are scaled equations for a flat torus)

Let r1^2 = x^2+y^2, r2^2 = w^2+z^2be the . (These are radial coordinates of two cylindrical coordinate systems filling out 4-space4-space). Then the norm squared is:

(1/2)*(r1^2+r2^2 + \sqrt{ (r1^2+r2^2)^2 - (r1^2-r2^2)^2 })


When this is less than one, this corresponds to the region plotted below:

We can now integrate over the shaded in region, \int_{region} dw dx dy dz.

But this

This 4-D integral can be reduced to 2D using r1 and r2, since dxdy dx dy = 2π r1 dr1, dw dz=2π dz = r2 dr2, and furthermore we can write r2 in terms of r1:

(4π^2) \int_{region} dr1 dr2 r1 r2


The region

Now, note that we can be rewrite r2 in terms of r1. In particular, after some manipulation of our norm, the shaded in region is defined by r2^2 ≤ 2-2\sqrt{2}r1+r1^2=(\sqrt{2}-r1)^2. Hence r2≤ \sqrt{2}-r1, and we can evaluate the r2 integral:

(4π^2) \int_{r1=0}^\sqrt{2} dr1 r1 \int_{r2=0}^{\sqrt{2}-r1} r2 dr2
= (4π^2) \int_{r1=0}^\sqrt{2} dr1 r1 (\sqrt{2}-r1})^2/2
= (4π^2) (1/6)


This then yields 2π^2/3, as Armin found.

2 more explicit calculation

I worked out the answer for the 2 by 2 case as well.

First, when dealing with 2 by 2 matrices in general, a convenient variable change is:

a->(w+x)/\sqrt{2},d->(w-x)/\sqrt{2},c->(y-z)/\sqrt{2},b->(y+z)/\sqrt{2}.


Then a^2+b^2+c^2+d^2 = w^2+x^2+y^2+z^2. And the determinant (ad-bc) = (1/2)*(x^2+y^2-w^2-z^2).

(Aside: this set of coordinates lets you see for instance that the set of rank 1 matrices in the space of 2D matrices realized as R^4 is a cone over the Clifford torus, since x^2+y^2 = w^2+z^2 on a sphere x^2+y^2+w^2+z^2=R^2 implies x^2+y^2 = R^2/2 and w^2+z^2 = R^2/2, which are scaled equations for a flat torus)

Let r1^2 = x^2+y^2, r2^2 = w^2+z^2 be the radial coordinates of two cylindrical coordinate systems filling out 4-space. Then the norm squared is:

(1/2)*(r1^2+r2^2 + \sqrt{ (r1^2+r2^2)^2 - (r1^2-r2^2)^2 })


This

When this is less than one, this corresponds to the region plotted below:

We can now integrate over the shaded in region, \int_{region} dw dx dy dz.

But this 4-D integral can be reduced to 2D using r1 and r2, since dxdy = 2π r1 dr1, dw dz=2π r2 dr2, and furthermore we can write r2 in terms of r1:

(4π^2) \int int_{region} dr1 dr2 r1 r2


with r2^2=2-2\sqrt{2}r1+r1^2, and the integrals running over positive r1,r2, obviously

The region can be defined by r2^2 2-2\sqrt{2}r1+r1^2=(\sqrt{2}-r1)^2.

With the substitution for r2, Hence r2≤ \sqrt{2}-r1, and we can make evaluate the above integral into a 1D r2 integral, with all the right substitutions. :

(4π^2) \int_{r1=0}^\sqrt{2} dr1 r1 \int_{r2=0}^{\sqrt{2}-r1} r2 dr2
= (4π^2) \int_{r1=0}^\sqrt{2} dr1 r1 (\sqrt{2}-r1})^2/2
= (4π^2) (1/6)


This then yields 2π^2/3, as Armin found.

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