Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am currently reading Quantum Invariants of Knots and 3-Manifolds by Turaev, and I am having a hard time understanding a statement made on page 120. He is explaining the property of space-structures, functors $U$ between some category of topological spaces and their homeomorphisms to the category of sets and bijections. He defines a structure as involutive if

  1. the canonical mapping $U(X) \times U(Y) \rightarrow U(X\sqcup Y)$ is equivariant (where $X$ and $Y$ are topological spaces and $\sqcup$ is a disjoint union)
  2. for any homeomorphism $f$ between spaces $X$ and $Y$, the induced bijection $U(f)$ is equivariant

He then goes on to what he considers an "obvious" example of the category of n-dimensional topological manifolds where $U$ takes $X$ to the set of all smooth structures on $X$, and claims that this functor is involutive without much explanation. Why is this example of a space structure so clearly involutive? To what extent is the above definition of involutive similar to the traditional one of $f(f(X))=X$?

share|cite|improve this question

You're missing part of the definition. A space structure $\mathfrak{A}$ is involutive if for every topological space $X$, there is an involution $i_X: \mathfrak{A}(X) \longrightarrow \mathfrak{A}$ satisfying the following properties:

  1. The canonical mappings giving compatibility with disjoint unions are equivariant with respect to the involutions. If we write $c_{X,Y}: \mathfrak{A}(X) \times \mathfrak{A}(Y) \longrightarrow \mathfrak{A}(X \amalg Y)$ for the canonical map associated to $X$ and $Y$, this means that $$c_{X,Y} \circ (i_X \times i_Y) = i_{X \amalg Y} \circ c_{X,Y}.$$

  2. For any homeomorphism $f: X \longrightarrow Y$, the induced bijection $\mathfrak{A}(f): \mathfrak{A}(X) \longrightarrow \mathfrak{A}(Y)$ is equivariant with respect to the involutions. This means that $$\mathfrak{A}(f) \circ i_X = i_Y \circ \mathfrak{A}(f).$$

The reason his example is trivial is because he takes $i_X = \mathrm{Id}_{\mathfrak{A}(X)}$ for all $X$. The basic nontrivial example you should keep in mind is oriented smooth manifolds, with the involution being orientation reversal. In fact, most other TQFT books just deal with the case of oriented cobordisms between smooth manifolds. Turaev's book tends to give constructions is greater generality than you will tend to see in practice.

share|cite|improve this answer
Interesting, do you have a recommendation for a reference that covers the topic in a similar rigor but restricted to oriented cobordisms between smooth manifold? – Benjamin Horowitz Jan 31 '14 at 5:14
There's nothing particularly wrong with Turaev's book. You can read the original Reshetikhin-Turaev papers to see their construction of a 3-manifold invariant. The papers are very readable, especially if you have already been going through Turaev's book. Another reference is Kock's book, but it only treats the case of 2D TQFTs. Kock's book is very easy to read, however, and will give you a good idea of what TQFT is about. – Henry T. Horton Jan 31 '14 at 15:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.