Let be a -graded vector space, where both and are finite-dimensional. Suppose that is a degree-preserving endomorphism of (so it consists of endomorphisms for both ). Then the *supertrace* of is defined as the difference of the two traces:

Obviously, the definition of the supertrace is very simple, but it begs for a justification. For example, why is this better than just adding the two traces? I’ve seen two facts offered in defense of the supertrace (see the Wikipedia article): 1) The supertrace is invariant when applied to endomorphisms of modules over commutative superalgebras, and 2) The supertrace of a supercommutator is 0.

I feel like neither of these justifications is particularly satisfying; the first requires a much more complicated setup than what I described above, and the second passes the burden of proof onto the notion of supercommutator. So I’d like to offer another way of thinking about the supertrace, which should be reasonably convincing to anyone who knows a little bit of homological algebra. I feel like this is one of those ideas that the practitioners in the field all know, but nobody bothers to write down.

Suppose that is actually a chain complex. This means that it has an odd operator such that . If you prefer, you could imagine a bounded -graded chain complex, where all the even-degree parts are combined to make , and all the odd-degree parts are combined to make .

As usual, you can define the homology . If is a chain map from to itself, then it induces an endomorphism of the homology. It’s a fairly easy exercise to show that . So the point is that the supertrace gives you information about how acts on the homology (or, more precisely, what the homology would be if you had a differential that was compatible with ), without requiring you to actually compute the homology. If you’re comfortable with this language, it may be better to say that the supertrace gives you homotopy-invariant information about .

For example, the supertrace of the identity map is just the Euler characteristic , which is well-known to be equal to , *regardless of what is*. So you could think of the supertrace as being a generalization of the Euler characteristic in the presence of an endomorphism.

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