The Unapologetic Mathematician

Mathematics for the interested outsider

The Determinant of Unitary and Orthogonal Transformations

Okay, we’ve got groups of unitary and orthogonal transformations (and the latter we can generalize to groups of matrices over arbitrary fields. These are defined by certain relations involving transformations and their adjoints (transposes of matrices over more general fields). So now that we’ve got information about how the determinant and the adjoint interact, we can see what happens when we restrict the determinant homomorphism to these subgroups of \mathrm{GL}(V).

First the orthogonal groups. This covers orthogonality with respect to general (nondegenerate) forms on an inner product space \mathrm{O}(V,B), the special case of orthogonality with respect to the underlying inner product \mathrm{O}(V), and the orthogonal matrix group over arbitrary fields \mathrm{O}(n,\mathbb{F})\subseteq\mathrm{GL}(n,\mathbb{F}). The general form describing all of these cases is

\displaystyle O^*BO=B

where O^* is the adjoint or the matrix transpose, as appropriate. Now we can take the determinant of both sides of this equation, using the fact that the determinant is a homomorphism. We find

\displaystyle\det(O^*)\det(B)\det(O)=\det(B)

Next we can use the fact that \det(O^*)=\det(O). We can also divide out by \det(B), since we know that B is invertible, and so its determinant is nonzero. We’re left with the observation that

\displaystyle\det(O)^2=1

And thus that the determinant of an orthogonal transformation O must be a square root of {1} in our field. For both real and complex matrices, this says \det(O)=\pm1, landing in the “sign group” (which is isomorphic to \mathbb{Z}_2).

What about unitary transformations? Here we just look at the unitarity condition

\displaystyle U^*U=I_V

We take determinants

\displaystyle\det(U^*)\det(U)=\det(I_V)=1

and use the fact that the determinant of the adjoint is the conjugate of the determinant

\displaystyle\overline{\det(U)}\det(U)=\lvert\det(U)\rvert^2=1

So the determinant of a unitary transformation U must be a unit complex number in the circle group (which, incidentally, contains the sign group above).

It seems, then, that when we take determinants the analogy we’ve been pushing starts to come out. Unitary (and orthogonal) transformations are like complex numbers on the unit circle, and their determinants actually are complex numbers on the unit circle.

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July 31, 2009 - Posted by | Algebra, Linear Algebra

2 Comments »

  1. […] the determinant of a unitary transformation is a unit complex number, and the determinant of a positive-semidefinite transformation is a nonnegative real number. If is […]

    Pingback by Polar Decomposition « The Unapologetic Mathematician | August 19, 2009 | Reply

  2. […] the determinant on itself, but we can easily restrict it to any subgroup. We actually know that for unitary and orthogonal transformations the image of this homomorphism must lie in a particular subgroup of . But in any case, the […]

    Pingback by The Special Linear Group (and others) « The Unapologetic Mathematician | September 8, 2009 | Reply


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