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Let $C$ be an algebra. Let $E = C^{\otimes 2n}$ be the tensor product (over the ground field) of $2n$ copies of $C$. [EDIT: Or better, $E = C\otimes C^{op}\otimes C\otimes C^{op}\cdots\otimes C \otimes C^{op}\quad$ ($2n$ copies of $C$, half of them "op").]

Let $F_{even}$ be the $E$ module with underlying vector space $C^{\otimes n}$, where the $(2i)$-th factor of $E$ acts on the left side of the $i$-th factor of $F_{even}$, and the $(2i+1)$-th factor of $E$ acts on the right side (i.e. left action of $C^{op}$) of the $i$-th factor of $F_{even}$. ($0 \le i \le n-1$ .)

Define $F_{odd}$ similarly (same underlying vector space $C^{\otimes n}$), but with the $(2i)$-th factor of $E$ acting on the left side of the $i$-th factor of $F_{odd}$, and the $(2i+1)$-th factor of $E$ acting on the right side of the $(i+1)$-th (modulo $n$) factor of $F_{odd}$. (Note that "modulo $n$" means the $n$-th factor is the 0-th factor.)

I'm interested in the derived $Hom_E(F_{even}, F_{odd})$. We can think of this as living in a disk with its boundary divided into $2n$ segments, alternating between incoming and outgoing segments. For $n=1$ this is just the Hochschild cohomology of $C$. For $n=2$ it's what one might associate to a saddle bordism in a 1+1-dimensional TQFT, or rather, the space in which the TQFT invariant of the saddle bordism lives.

Does $Hom_E(F_{even}, F_{odd})$ have a name? Is it mentioned in the literature anywhere?

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I'm confused; don't you need some of the copies of C in E to be opposite algebras, or have an involution, in order to have it act on the right-hand sides? –  Tyler Lawson Dec 9 '09 at 2:48
    
Yes, good point. I'll edit the question to clarify this. –  Kevin Walker Dec 9 '09 at 3:27
    
In what you just edited you have a direct product of algebras where it should be tensor product. –  Grétar Amazeen Dec 9 '09 at 4:22
    
Now I'm confused. I don't think you need to involve the opposite algebra, since it makes perfect sense to let C act on C from the right. In general you would need to have the opposite algebra in there, but C is naturally a right C-module as well as a left C-module. Another thing, if you change it to C \otimes C^op \otimes... etc, then the definition of F_even doesn't really make sense, does it? Now you have the opposite algebra acting from the right, which is the same as the original algebra acting from the left, so for the i-th guy in C^{\otimes n} you have two copies of C acting on the left. –  Grétar Amazeen Dec 9 '09 at 4:54
    
And similarly for F_odd. –  Grétar Amazeen Dec 9 '09 at 4:54
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