## The Center of an Algebra

Sorry I forgot to get this posted this morning.

Given an algebra , it’s interesting to consider the “center” of . This is the collection of algebra elements that commute with all the others. That is,

It’s straightforward to see that sums, scalar multiples, and products of central elements — elements of — are themselves central. That is, is an algebra, and it’s a commutative one to boot. This gives us a construction that starts with an associative algebra and ends with a commutative algebra, and yet it turns out that it is *not* a functor! I don’t really want to get into that right now, though, but I wanted to mention it in passing, since it’s one of the few examples of a natural algebraic construction that isn’t functorial.

What I do want to get into right now, is calculating the center of the matrix algebra . The answer is reminiscent of Schur’s lemma:

Suppose that is a central matrix. Then in particular it commutes with the matrix , which has a at the th place along the diagonal and s everywhere else. That is, . But zeroes out everything except the th column of , while zeroes out everything except the th row. For these two be equal, the th column must be all zeroes except for the one spot along the diagonal, and similarly for the th row. And so must be diagonal.

For , must also commute with — the matrix with ones in the th column of the th row and the th column of the th row. That is, . Multiplying on the right by swaps the th and th columns of , while multiplying on the left swaps the th and th rows. Thus we can tell that not only is diagonal, but all the diagonal entries must be the same. And so for some complex .

[…] is only possible if for each we have for all . But this means that is in the center of , which implies that . Therefore a central element can be […]

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[…] which goes summand-by-summand. Each summand is (isomorphic to) a complete matrix algebra, so we know that its center is isomorphic to . Thus we find that the center of is the direct sum of copies of […]

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