Homotopy Fixed Points of SO(2) on Fully Dualizable Algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T03:34:06Z http://mathoverflow.net/feeds/question/73922 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73922/homotopy-fixed-points-of-so2-on-fully-dualizable-algebras Homotopy Fixed Points of SO(2) on Fully Dualizable Algebras Sam Gunningham 2011-08-28T22:33:59Z 2011-08-29T16:45:16Z <p>Note: by fixed points, I always mean homotopy fixed points.</p> <p>As explained in Jacob Lurie's <a href="http://www.math.harvard.edu/~lurie/papers/cobordism.pdf" rel="nofollow">paper</a> on the cobordism hypothesis, we have an action of O(2) on the $\infty$-groupoid $X$ given by considering fully dualizable objects and invertable morphisms in some symmetric monoidal $(\infty ,2)$ category $\mathcal C$. I am interested in the case when $\mathcal C$ is the category where objects are algebras, 1-cells are bimodules, 2-cells maps of bimodules etc... By an algebra, I want to include algebra objects in some $\infty$-category (e.g. chain complexes), so that $\mathcal C$ really has non-trivial 3-cells, 4-cells and so on.</p> <p>Now, I know (from remark 4.2.7) that the fixed points for the induced action of SO(2) on this space correspond to cyclic Frobenius algebras, i.e. smooth, proper algebras $A$, with a non-degenerate <em>and $SO(2)$-equivariant</em> trace</p> <p>$A \otimes _{A^e} A \to k$.</p> <p>Here, non-degenerate means that the pairing $A\otimes A \to A\otimes _{A^e} A \to k$ is nondegenerate. Recall also that the Hochschild homology $A \otimes _{A^e} A$ carries and $SO(2)$ action, so it makes sense to talk about $SO(2)$-equivariant map above. </p> <p>The data of SO(2) equivariance should be equivalent to descending the trace to cyclic homology.</p> <p>However, the reason why the fixed points are as above is that both can be identified with 2-d oriented TFTs, after theorem 3.1.8 which describes how to extend from a (n-1)-dimensional TFT to an n-dimensional one.</p> <p><strong>Question: Can we calculate this action and identify the fixed point space directly?</strong></p> <p>Part of this is not too hard to see (I think): naively, the SO(2) action on $X$ gives a canonical (Morita) automorphism of each (f.d.) algebra $A$. This automorphism is given by the bimodule dual $A^!$ (or the other one $A^\vee$...), so being a fixed point means to give an isomorphism $A^! \to A$, which is the same as a non-degenerate map $A\otimes _{A^e} A \to k$. If the base category where our algebra lives has no higher structure, then I think this is enough, but this does not explain the $SO(2)$ equivariance.</p> <p><strong>In particular, where does the SO(2)-equivariance data come from?</strong></p> <p>To see this, I tried to go a little deeper (with help from Takuo Matsuoka): the action is given by a map $B\mathbb Z =SO(2) \to Aut(X)$, which we transformed into a map $\mathbb Z \to \Omega Aut(X)$. The data described above corresponds to picking an algebra $A$ and composing to get a map $\mathbb Z \to \Omega X$ = Invertable $A-A$-bimodules. However, we haven't used that the original map from SO(2) is a group homomorphism - this should correspond to the map $\mathbb Z \to \Omega Aut(X)$ being an $E_2$-map. So we get braiding data attached to each of the $A^!$, which should be trivialized (because the E_2 -structure on $\mathbb Z$ is trivial (?)). Then to give a fixed point we must take into account all this data...</p> <p>This has only given me a headache so far, but maybe there is an easier way to think about this? I tried (failed?) to be brief, so let me know if anything here is unclear!</p> http://mathoverflow.net/questions/73922/homotopy-fixed-points-of-so2-on-fully-dualizable-algebras/73965#73965 Answer by Noah Snyder for Homotopy Fixed Points of SO(2) on Fully Dualizable Algebras Noah Snyder 2011-08-29T12:38:40Z 2011-08-29T12:38:40Z <p>I'm not comfortable enough with the $\infty$ setting to give a full answer, but for ordinary algebras the condition you're looking for here should be that of a <em>symmetric</em> Frobenius algebra (i.e. the Frobenius bilinear form is symmetric). So you're just looking for the $\infty$ analogue of the symmetric condition on Frobenius algebras.</p> <p>There's a lot of great information about oriented 2d extended TFTs in <a href="http://sites.google.com/site/chrisschommerpriesmath/" rel="nofollow">Chris Schommer-Pries's thesis</a>.</p> http://mathoverflow.net/questions/73922/homotopy-fixed-points-of-so2-on-fully-dualizable-algebras/73986#73986 Answer by Chris Schommer-Pries for Homotopy Fixed Points of SO(2) on Fully Dualizable Algebras Chris Schommer-Pries 2011-08-29T16:45:16Z 2011-08-29T16:45:16Z <p>I might be confused about your question. Are you asking...</p> <ol> <li>How is trivializing the $O(n)$-action the same as giving an $O(n)$-equivariant non-degenerate trace? (as per Lurie's theorem 3.1.8).</li> <li>How can we identify the $SO(2)$-action with the usual $SO(2)$-action on Hochschild homology?</li> </ol> <p>For the first one, I think this is pretty well described in Jacob's paper. The key is understanding what the condition "non-degenerate" means in that setting. This is the condition that allows you to turn the equivariant trace into the trivialization of an action. </p> <p>If it is the second thing you are asking, here is one way to see this, though there may be an easier way. Let's first discuss the more direct approach which is giving you a headache. Instead of thinking about $E_2$-structures or (homtopical) group homomorphisms, I prefer to think of the map of delooings:</p> <p>$$\alpha: BSO(2) = K(\mathbb{Z}, 2) = \mathbb{CP}^\infty \to BAut(X)$$</p> <p>This is equivalent to understanding the $E_2$-structure you mentioned, but for me it is easier to comprehend. A (homotopical) $SO(2)$-action on $X$ is the same as the map $\alpha$. I find this easier probably because of my background. Now $BSO(2) = \mathbb{CP}^\infty$ has a well known cell structure which goes:</p> <p>$$S^2 \cup e_4 \cup e_6 \cup \dots$$</p> <p>So part of the $SO(2)$-action is an element in $\pi_2 BAut(X)$ which gives the invertible $A-A$-bimodule you identified up above. This is also called the Serre bimodule in this context. The element in $\pi_2$ actually contains more information: it is a <strong>natural transformation</strong> (from the identity functor to itself). </p> <p>From this point of view, to understand the rest of the $SO(2)$-action, you need to extend this map all the way through the entire CW-structure. You can start doing this explicitly. For example the fact that the Serre is a natural transformation allows you to construct an element in $\pi_3 BAut(X)$ which is essentially the "self braiding" of the Serre with itself. Trivializing this element allows you to extend the map to the 4-skeleton $\mathbb{CP}^2 = S^2 \cup e_4$. This then gives rise to a higher "self braiding" which you then have to trivialize, etc. </p> <p>This quickly becomes a huge headache, as each of thee trivializations is supposed to be describable in terms of the basic fully-dualizable structure. While possible, it is not entirely obvious how to procede at each step. Also, even if we succeed this doesn't yet do what you want; we still haven't identified this SO(2)-action with the usual one on Hochschild homology. </p> <p>A different approach is to realize that $$BSO(2) \simeq |N \Lambda^{op}|$$ where $\Lambda$ is the cyclic category. This allows us to get a different "filtration" of $K(\mathbb{Z}, 2)$. If we view the map $\alpha$ as classifying an $X$-bundle over $BSO(2)$, then we can use this description, plus the material of Dwyer-Hopkins-Kan "The homotopy theory of cyclic sets" to realize that this is equivalent to the data of a certain cyclic set. </p> <p>I think that in the universal case we get a cyclic set which essentially comes from the various ways to decompose a circle cyclicly, and in the specific case at hand we recover the Hochschild homology complex. This would identify the two circle actions. I hope this helps. </p>