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Chas and Sullivan constructed in 1999 a Batalin-Vilkovisky algebra structure on the shifted homology of the loop space of a manifold: $\mathbb{H}_*(LM) := H_{*+d}(LM;\mathbb{Q})$. This structure includes a product which combines the intersection product and Pontryagin product and a BV operater $\Delta: \mathbb{H}_*(LM) \to \mathbb{H}_{*+1}(LM)$.

I was wondering about the applications of this structure. Has it even been used to prove theorems in other parts of mathematics? A more concrete question is the following: Usually, considering a more complicated structure on topological invariants of a space allows you to prove certain non-existince results. For example, the cup product in cohomology allows you to distinguish between $ S^2 \vee S^1 \vee S^1 $
and $T^2$. Is there an example of this type for string topology?

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Hossein Abbaspour gave an interesting connection between 3-manifold topology and the string topology algebraic structure in arXiv:0310112. The map $M \to LM$ given by sending a point $x$ to the constant loop at $x$ allows one to split

$\mathbb{H}_*(LM)$ as $H_*(M) \oplus A_M$.

He showed essentially that the restriction of the string product to the $A_M$ summand is nontrivial if and only if $M$ is hyperbolic. There are some technical details in the statements in his paper, but it was written pre-Perelman and I believe the statements can be made a bit more elegant in light of the Geometrization Theorem.

Philosophically, Sullivan has said that he his goal in inventing string topology was to try to find new invariants of smooth structures on manifolds. His original idea was that if you have to use the smooth structure to smoothly put chains into transversal positions to intersect them then you might hope that the answer will depend on the smooth structure. Unfortunately, we now know that the string topology BV algebra depends only on the underlying homotopy type of the manifold (there are now quite a few different proofs of various parts of this statement).

The string topology BV algebra is only a piece of a potentially much richer algebraic structure. Roughly speaking, $\mathbb{H}_*(LM)$ is a homological conformal field theory. This was believed to be true for quite some time but it took a while before it was finally produced by Veronique Godin arxiv:0711.4859. She constructed an action of the PROP made from the homology of moduli spaces of Riemann surfaces with boundary. Restricting this action to pairs of pants recovers the original Chas-Sullivan structure.

Unfortunately, for degree reasons, nearly all of the higher operations vanish. In particular, any operation given by a class in the Harer stable range of the homology of the moduli space must act by zero. Hirotaka Tamanoi has a paper that spells out the details, but it is nothing deep.

Furthermore, it seems that the higher operations are homotopy invariant as well. For instance Lurie gets this as a corollary of his work on the classification of topological field theories.

Last I heard, Sullivan, ever the optimist, believes that there is still hope for string topology to detect smooth structures. He says that one should be able to extend from the moduli spaces of Riemann surfaces to a certain piece of the boundary of the Deligne-Mumford compactification. I've heard that the partial compactification here is meant to be that one allows nodes to form, but only so long as the nodes collectively do not separate the incoming boundary components from the outgoing boundary. Sullivan now has some reasons to hope that operations coming from homology classes related to the boundary of these moduli spaces might see some information about the underlying smooth structure of the manifold.

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  • $\begingroup$ My two cents worth: one of the original motivations of Sullivan in creating string topology was the hope that the string topology operations would detect the smooth structure of the manifold. We now know this to be false (the homology level operations are oriented homotopy invariant). The last time I spoke with Dennis about this, he still held out the hope that the chain level operations would detect the smooth structure. But, it seems to me that Lurie's work (if I properly understand it) implies that the chain level operations are also homotopy invariant. $\endgroup$
    – John Klein
    Commented Jan 24, 2011 at 3:51
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    $\begingroup$ This is fascinating! Perhaps it's worth clarifying the interaction with geometrisation. From a look at Abbaspour's introduction, his main theorem characterises when $M$ is algebraically hyperbolic, i.e. when $M$ is aspherical and $\pi_1(M)$ contains no $\mathbb{Z}^2$ subgroup. Perelman's hyperbolisation theorem says that algebaically hyperbolic implies hyperbolic. In summary, it looks like Abbaspour's result is purely topological, and the content of the implied geometric statement is precisely hyperbolisation. $\endgroup$
    – HJRW
    Commented Jan 3, 2022 at 10:00
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As a shameless plug, I may say that in my thesis we do show that string topology, interpreted in a broader context, is NOT a homotopy invariant. What we do is the following : instead of looking at loops in $M$ we think of them as arcs in $M\times M$ with its boundary in the diagonal $M$ that sits inside $M\times M$. Now we look at the space of such arcs $\mathcal{S}(M)$, which, when they intersect the diagonal at intermediate stages, do so transversely. One can then define a suitable coalgebra structure which is NOT a homotopy invariant. In particular, this structure distinguishes the Lens spaces $L(7,1)$ from $L(7,2)$, which are homotopy equivalent but NOT homeomorphic.

Of course, this new structure is not related to the loop product or the BV operator as per the question asked. Moreover, this structure is defined on a much smaller space then $LM$. However, if you take the point of view that string topology is broadly the study of loops in a manifold then this is a new and interesting algebraic structure.

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Let's consider $SU(3)$ and $S^3\times S^5$. These two $8$-manifolds are not homotopy equivalent. You can't see this by looking at their cohomology groups or cohomology rings, but you can see it using the action of the Steenrod algebra: $Sq^2\colon H^3(M;\mathbb{Z}/2\mathbb{Z})\to H^5(M;\mathbb{Z}/2\mathbb{Z})$ is zero for $M=S^3\times S^5$ and nonzero for $M=SU(3)$.

We could also use string topology to distinguish between $SU(3)$ and $S^3\times S^5$. Here's how. First, look up the Batalin-Vilkovisky algebras. Tamanoi computed the result for $SU(n)$ and Menichi computed the results for spheres. It's enough to use $\mathbb{Z}/2\mathbb{Z}$ coefficients. Second, compare the two. You can see that they are isomorphic (boo) but that no such isomorphism would preserve the "constant loop summand" $\mathbb{H}_\ast(M)$ that sits inside the BV algebra. So in particular the isomorphism cannot come from a homotopy equivalence between the two manifolds.

Of course we already knew that these spaces were not homotopy equivalent, and it would have been much nicer if the BV algebras were not isomorphic at all. In general, computing the BV algebras is rather difficult, and it's probably not an efficient way to distinguish between manifolds.

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    $\begingroup$ The example you have mentioned works, in the general setting and the string topology setting, as there is a rational homotopy equivalence between $S^3\times S^5$ and $SU(3)$ but no homotopy equivalence. The reason that the fundamental class (the identity of the loop product) is not preserved is basically because that the best possible map $f:S^3\times S^5\to SU(3)$ maps $[S^3]$ to a generator of $H_3(SU(3))$ but $[S^5]$ gets mapped to twice a generator of $H_5(SU(3)$. $\endgroup$ Commented Mar 5, 2012 at 3:04
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Kallel and Salvatore use the string product to help compute the homology of a mapping space in this paper.

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The string topology of a manifold is isomorphic to the Hamiltonian Floer Homology of the cotangent bundle of the manifold.

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Even though one may argue that it is not stepping too much outside the area, in a sense, you might want to look at the paper of Xiaojun Chen, Wee Liang Gan, https://arxiv.org/abs/0804.4748.

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  • $\begingroup$ Neat paper! I don't know why I hadn't looked at it before now. $\endgroup$
    – GS
    Commented May 12, 2010 at 21:21

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