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Let $A=C^{\infty}(\mathbb{S}^{1})$, let $B$ be the sub-algebra $C^{\infty}(0,1)$. Here we identify $\mathbb{S}^{1}$ by $\mathbb{R}/2\pi \mathbb{Z}$. I want to ask if there is a way I can decompose $A$ to calculate $Tor_{B\otimes B^{op}}(B,A)$. I suspect it is the same as $Tor_{B\otimes B^{op}}(B,B)=HH(B)\cong H_{DR}((0,1))$, but I do not know a clean way to do it. Or maybe the conjectured relationship is false.

I should note that the isomorphism is somewhat unclear at chain level, since for projective tensor products we should have $C^{\infty}(M_1)\hat{\otimes}C^{\infty}(M_2)\cong C^{\infty}(M_1\times M_2)$ (I do not know a proof myself). And it is hard to believe $C^{\infty}(B\times\cdots B\times A)$ can be deformed into $C^{\infty}(B\times\cdots B\times B)$. Intuitively, $A$ contains "extra" info of functions smooth outside of $(0,1)$, and that should be a flat module over $B\otimes B^{op}$. But I do not know how to properly quanitfy it.

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  • $\begingroup$ Is this just "usual" Tor, or some version that takes into account topologies? $\endgroup$
    – Yemon Choi
    Commented May 21, 2016 at 3:05
  • $\begingroup$ @YemonChoi: I know $C^{\infty}(M)$ is in general a nuclear space and the "projective tensor product" is well defined and unique. However I cannot find any clear reference for a proof of this (seemingly trivial) fact. For your question, it is just the usual Tor (i.e, what you get by tensoring a projective resolution of B by itself). $\endgroup$
    – Bombyx mori
    Commented May 21, 2016 at 3:08
  • $\begingroup$ @YemonChoi: It is embarassing, but the reference online I can find on nuclear spaces usually goes back to Gronthedieck's original papers. And I cannot read French. So to me it is more of a black box. $\endgroup$
    – Bombyx mori
    Commented May 21, 2016 at 3:10
  • $\begingroup$ What category are you working in, and with what exact structure? Are you considering categories where objects have some topology and morphisms are required to be continuous? Or are you just considering these spaces as modules over a ring and allowing morphisms that need not be continuous? IIRC this affects what one means by projective, injective etc and hence could change one's Ext and Tor. $\endgroup$
    – Yemon Choi
    Commented May 21, 2016 at 3:52
  • $\begingroup$ @YemonChoi: I do need $C^{\infty}(M)$ to be a Frechet space with the locally convex topology induced by semi-norm. The morphisms should be continuous as you say. I think maybe the category of Frechet spaces is appropriate, since $C^{\infty}(M)$ is not a Banach space. $\endgroup$
    – Bombyx mori
    Commented May 21, 2016 at 4:05

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