Cobordism Hypothesis in dimension 1

Question

I am trying to understand consequences of the Cobordism Hypothesis in dimension 1, following section 4.2 of Lurie's "On the Classification of Topological Field Theories". Especially, I want to understand the '$S^1$ action' on the dimension of a dualizable object.

Given a symmetric monoidal $(\infty, 1)$-category $\mathcal{C}$ and a dualizable object $X\in \mathcal{C}$, by the Cobordism Hypothesis there exists a symmmetric monoidal functor $Z:\mathbf{Bord}^{\mathrm{or}}_1 \to \mathcal{C}$. Then the author asserts that $\mathrm{Map}_{\mathbf{Bord}^{\mathrm{or}}_1}(\emptyset, \emptyset)$ can be identified with a classifying space for oriented closed 1-manifolds, especially containing $BSO(2)\simeq \mathbb{CP}^\infty$ as a connected component. Then the map $f:\mathbb{CP}^\infty \to \mathrm{Map}_\mathcal{C}(1, 1)$ induced by $Z$ is supposed to encode this $S^1$ action on $\dim(X):1\to 1$.

I am confused about a lot of things in the paragraph explaining this. My questions are:

1. Why is $\mathrm{Map}_{\mathbf{Bord}^{\mathrm{or}}_1}(\emptyset, \emptyset)$ identified as a classifying space for oriented closed 1-manifolds? (and what does this classifying space mean exactly?)
2. What does the map $f:\mathbb{CP}^\infty \to \mathrm{Map}_\mathcal{C}(1, 1)$ have to do with a $S^1$ action?
3. What does this '$S^1$ action' really mean exactly? What kind of object is this $S^1$, and what does it mean for this to act on an object in a category?

Answer 1

Question 1. In general and informally, given a type of geometric objects (e.g. smooth curves), a classifying space for that type of objects should be a space $\mathcal{M}$ with the property that a map $*\to\mathcal{M}$ should correspond to one of those geometric objects (e.g. a smooth curve), and more generally a map $B\to\mathcal{M}$ should correspond to a family of those geometric objects living over $B$ (e.g. a family of smooth curves living over $B$). In thise sense, for instance, the classifying space $BG$ of a group classifies $G$-torsors, if you set things up in a proper $(\infty,1)$-categorical way. Now, a smooth oriented $1$-bordism from the empty manifold $\varnothing$ to itself is exactly the data of a smooth oriented closed $1$-manifold, so the points of the mapping space $\mathrm{map}_{\mathbf{Bord}^\mathrm{or}_1}(\varnothing,\varnothing)$ are those (smooth) oriented closed $1$-manifolds, i.e. a map $*\to\mathrm{map}_{\mathbf{Bord}^\mathrm{or}_1}(\varnothing,\varnothing)$ corresponds to some oriented closed $1$-manifold. You can likewise convince yourself that, for any homotopy type $B$, a map $B\to\mathrm{map}_{\mathbf{Bord}^\mathrm{or}_1}(\varnothing,\varnothing)$ corresponds to some notion of ''a family of oriented closed $1$-manifolds over the homotopy type $B$''. In this sense, $\mathrm{map}_{\mathbf{Bord}^\mathrm{or}_1}(\varnothing,\varnothing)$ gives you a classifying space of oriented closed $1$-manifolds.

Question 3. (We will address question 2. momentarily.) Given an $(\infty,1)$-category $\mathscr{C}$ with an object $X$ in it and a group $G$, we can let an action of $G$ on $X$ be an $\mathbb{E}_1$-group homomorphism $G\to\mathscr{C}^\simeq(X,X)$, where $\mathscr{C}^\simeq$ is the core of $\mathscr{C}$. Here, an $\mathbb{E}_1$-group is a homotopy coherent analogue of a group inside the $(\infty,1)$-category of spaces. It is a fact that, for any $\infty$-groupoid (such as $\mathscr{C}^\simeq$), the endomorphism spaces carry such a structure of an $\mathbb{E}_1$-group with composition as group law. Moreover, we also used that any ordinary (or topological) group has an underlying $\mathbb{E}_1$-group structure, which for ordinary groups in particular is considering a set as a discrete space. Note that our definition of a group action is a generalization of the idea that an action of a group $G$ on some object $X$ with automorphism group $\mathrm{Aut}(X)$ should just be a group homomorphism $G\to\mathrm{Aut}(X)$. (Especially if $G$ and $X$ do not lie in the same category, you cannot really try to make the other kind of definition precise, namely that in which you write down a map $G\times X\to X$ satisfying a bunch of properties. We are sort of currying over a map $G\times X\to X$ to a map $G\to\mathrm{Aut}(X)$ and take that as starting point for our definition.)

Question 2. The circle $S^1$ is an oriented circle bundle, and hence corresponds to some map $\sigma\colon *\to\mathbb{CP}^\infty\simeq BSO(2)$, and we can consider $SO(2)\simeq S^1$. Hence $\sigma$ is a special case of the following: for any group $G$, the $(\infty,1)$-categorical definition of $BG$ as homotopy colimit gives a canonical map $*\to BG$. It holds that $G\simeq *\times_{BG} *$ (where we use this canonical map on both sides of the pullback). I have explained why in this answer here, at the end of the first paragraph of the proof of Lemma 2. Hence we have $S^1\simeq *\times_{\sigma,\mathbb{CP}^\infty,\sigma} *\simeq\mathrm{map}_{\mathbb{CP}^\infty}(\sigma(*),\sigma(*))$, where I am considering $\mathbb{CP}^\infty$ as an $\infty$-groupoid in the last step. By functoriality of $f$, therefore, the map $\mathbb{CP}^\infty\to\mathrm{map}_\mathcal{C}(\mathbf{1},\mathbf{1})$ hence induces a map $$ S^1\simeq\mathrm{map}_{\mathbb{CP}^\infty}(\sigma(*),\sigma(*))\to\mathrm{map}_{\mathrm{map}_\mathcal{C}(\mathbf{1},\mathbf{1})}(f\sigma(*),f\sigma(*)) $$ and now we use that $f\sigma(*)\simeq Z(S^1)$ where we consider $S^1\in\mathrm{map}_{\mathbf{Bord}^\mathrm{or}_1}(\varnothing,\varnothing)$. This holds because $f$ itself is induced by $Z$, and the composite map $\sigma\colon *\to\mathbb{CP}^\infty\hookrightarrow\mathrm{map}_{\mathbf{Bord}^\mathrm{or}_1}(\varnothing,\varnothing)$ sends $*$ to $S^1$.

This map $S^1\to\mathrm{map}_{\mathrm{map}_\mathcal{C}(\mathbf{1},\mathbf{1})}(Z(S^1),Z(S^1))$ is in fact an $\mathbb{E}_1$-group homomorphism. Since $f$ preserves composition as functor between $\infty$-groupoids, the only thing we need to convince ourselves of is that the equivalence $S^1\simeq\mathrm{map}_{\mathbb{CP}^\infty}(\sigma(*),\sigma(*))$ preserves the $\mathbb{E}_1$-group structure. I will not show this, but it sort of follows if you think long enough about how spaces and $\infty$-groupoids are the same, and what in spaces corresponds to composition in $\infty$-groupoids.

Answer 2

In the setting of ordinary category theory, the oriented bordism category $\mathrm{Bord}_1^\text{or}$ has objects given by disjoint unions of oriented points ($0$-manifolds) and morphisms given by the set of oriented bordisms between these collections of oriented points. In particular, the morphisms $\hom_{\mathrm{Bord}_1^\text{or}}(\emptyset, \emptyset)$ can be identified with the set of oriented closed $1$-manifolds, i.e., disjoint unions of oriented circles.

In the setting of higher category theory, the $(\infty, 1)$-category $\mathbf{Bord}_1^\text{or}$ still has objects given by disjoint unions of oriented points, but the morphisms between such collections is now a $(\infty, 0)$-category (i.e., an $\infty$-groupoid/space/anima) of bordisms. If $M, N$ are two oriented $0$-manifolds, then objects of the $(\infty, 0)$-category $\operatorname{map}_{\mathbf{Bord}_1^\text{or}}(M, N)$ are still just the set of $1$-bordisms between $M$ and $N$ as before, and now we have (invertible) morphisms in $\operatorname{map}_{\mathbf{Bord}_1^\text{or}}(M, N)$ given by orientation-preserving diffeomorphisms between these $1$-bordisms that restrict to the identity on $M$ and $N$.

Specializing to the case $M = N = \emptyset$, we see that $\operatorname{map}_{\mathbf{Bord}_1^\text{or}}(\emptyset, \emptyset)$ has objects given by the set of oriented closed $1$-manifolds, and morphisms given by orientation-preserving diffeomorphisms of these oriented closed $1$-manifolds. It is in this sense $\operatorname{map}_{\mathbf{Bord}_1^\text{or}}(\emptyset, \emptyset)$ is a classifying space for oriented closed $1$-manifolds. More explicitly, by restricting to a skeleton of $\operatorname{map}_{\mathbf{Bord}_1^\text{or}}(\emptyset, \emptyset)$, we have $$\operatorname{map}_{\mathbf{Bord}_1^\text{or}}(\emptyset, \emptyset) \simeq \bigsqcup_Z B\operatorname{Diff}^+(Z),$$ where $Z$ ranges over oriented diffeomorphism equivalence classes of oriented closed $1$-manifolds.

The path component of $\operatorname{map}_{\mathbf{Bord}_1^\text{or}}(\emptyset, \emptyset)$ corresponding to $Z = S^1$ is $$B\operatorname{Diff}^+(S^1) \simeq BSO(2) \simeq \mathbb{C}P^\infty.$$ Now, given a symmetric monoidal functor $Z: \mathbf{Bord}_1^\text{or} \to \mathbf{C}$, we have an induced map $f: \mathbb{C}P^\infty \subset \operatorname{map}_{\mathbf{Bord}_1^\text{or}}(\emptyset, \emptyset) \to \operatorname{map}_{\mathbf{C}}(1, 1)$. Recall that $\mathbb{C}P^\infty \simeq BS^1$ (since $SO(2) \cong S^1$). This means that the map $f$ equips $f(*) \in \operatorname{map}_{\mathbf{C}}(1, 1)$ with the action of the circle $S^1$. Thinking in terms of ($\infty, 0$)-categories again, $BS^1$ is the nerve of a category with one object $*$ whose automorphisms are given by the group $S^1$. Therefore, by functoriality these automorphisms give rise to automorphisms $f(*) \to f(*)$ indexed by $S^1$ that compose according to the group law in $S^1$.