Math 202B: Revenge of the Centralizer

Yesterday during office hours I had the unsettling experience of being shown that something I assumed to be true is, in fact, subtly incorrect. That’s what happens when you take things for granite. My sloppy thinking was pointed out to me by Amelia and Eliza in the context of the following problem which appeared on the Math 202B Exam: prove that if \mathcal{B} is a maximal abelian subalgebra of an algebra \mathcal{A}, then the centralizer Z(\mathcal{B},\mathcal{A}) of \mathcal{B} in \mathcal{A} equals \mathcal{B}.

This was supposed to be easy. Since \mathcal{B} is abelian, it is certainly contained in its centralizer Z(\mathcal{B},\mathcal{A}). Now, if the containment is proper, then there exists an element C \in Z(\mathcal{B},\mathcal{A}) which does not lie in \mathcal{B}. Then,

\mathcal{C}=\mathrm{CAlg}(\mathcal{B} \sqcup \{C\}),

the commutative algebra generated by \mathcal{B} with C adjoined, is a commutative algebra properly containing \mathcal{B}, contradicting the fact that \mathcal{B} is a MASA.

The subtle issue here is that \mathcal{C} is by definition the intersection of all commutative subalgebras of \mathcal{A} containing the set \mathcal{B} \sqcup \{C\}, and this could conceivably be empty. To be concrete, imagine a single non-normal matrix C. There is no commutative subalgebra of the full matrix algebra containing C, since our definition of “algebra” requires *-closure, and by hypothesis C does not commute with its adjoint.

Luckily, our attempted solution can be salvaged: as Evan immediately recognized, if we can show that Z(\mathcal{B},\mathcal{A}) contains a *selfadjoint* element D not contained in \mathcal{B} this issue goes away. This is the case, since we can take D to be either the real or imaginary part of C — at least one of these does not lie in \mathcal{B}, since if they both did so would C.

Hopefully I’m now getting this right, and my apologies for being such a bad professor. Try reading some of my papers, I swear they’re really good.

In any case, we will have to incorporate this correction into our census of reality. In particular, if this problem were to appear on the qual, a proposed solution which did not address this issue would no longer pass muster. Live and learn.

1 Comment

  1. Li says:

    Lovely post. Makes my head spin

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