4

When studying physics, I came up with various integrals that only takes integer values due to topological reasons. Winding number $S^1 \to S^1$ is the most elementary example, which moreover provides an isomorphism $\pi_1(S^1) \to \mathbb Z$. More generally, homotopy class of a map from $S^d$ or $T^d$ to a Lie group $G$ is often represented by an integral, as I saw in various physics literature (one example is Homotopy and quantization in condensed matter physics, written by mathematical physicists).

I want a systematic theory on this matter.

Already in this site, there are at least two questions related to my question, which focuses on a specific choice of $X$ and $Y$ when classifying maps $X\to Y$:

Why is the winding number of a matrix an integer?

Cartan 3-form on a Lie group G

What I want to address is more general, since I feel that there is a more general theory of obtaining homotopy invariant as an integral. Let $X = S^d$ or $T^d$, and let $G$ be a "nice" Lie group (say, compact).

  1. What is the homotopy class $[X, G]$?

  2. Can we associate an invariant written in terms of an integral over $X$ that takes only integer values?

  3. If such invariant exists, when does this invariant completely characterizes the homotopy class? In particular, in the link Why is the winding number of a matrix an integer? , a map $f:S^1→U(N)$ is classified by the degree of the map $\det\circ f:S^1\to S^1$. Then, does this invariant completely characterizes $[S^1,U(N)]$?

Any answers or references on this question is welcomed. I have a basic knowledge of algebraic topology and differential geometry.

Note added: After several comments, I can make my question more concrete. According to a comment:

If $f:X→G$ is smooth and $\omega$ is a $k$-form representing an integer cohomology class on $G$, then $f^∗\omega$ will be a $k$-form on $X$ representing an integer cohomology class. In particular, integrating it against an integer homology class will give an integer invariant of $f$.

My question following the comment is:

  1. How to find a $k$-form representing an integer cohomology class?
  2. What is the "integer homology class"? Is it correct to say that $\int_N f^*\omega \in \mathbb Z$ whenever $N$ is a $k$-dimensional submanifold of $M$?
Laplacian
  • 2,408
  • 2
    I lack enough knowledge in homotopy theory to properly address your very interesting questions. I would like to point out, if you haven't already encountered the theory before, to Chern-Weil theory, in which integrals of certain cohomology classes characterize topologically principal or vector bundles, spitting out, basically, integer numbers. – topolosaurus Jun 22 '22 at 11:42
  • @topolosaurus Thanks for your comment. I have a very basic knowledge of Chern character, learned from a differential geometry book by Tu (one of the GTM series book). In that book, given a vector bundle $E$ on $M$, the Chern class $c_i(E)$ is defined, but the "integrality" is not proved. However, my primary interest is that why $\int_M c_i(E)$ is an "integer". Could you suggest a book or lecture note on why such integral is an "integer"? – Laplacian Jun 22 '22 at 12:09
  • 1
    The Kronecker pairing between an integral cohomology class and the fundamental class is always an integer. So, why is the Chern class an integral cohomology class (up to appropriate scaling)? In some definitions, this is essentially a tautology, so this depends on the definition you're using. – JHF Jun 22 '22 at 15:29
  • 1
    Well, if you hope to understand all of $[X,G]$ in terms of integrals (at least using smooth tools like de Rham cohomology), the first step is going to need to be showing that every map in $[X,G]$ is homotopic to a smooth one. Since you're assuming $X$ is compact, this is true if $G$ is compact, and there may be Lie group tools that you can use more generally. – jgon Jun 22 '22 at 17:14
  • 2
    For 2, in general if $f:X\to G$ is smooth, $\omega$ is a $k$-form representing an integer cohomology class on $G$, then the pullback $f^*\omega$ will be a $k$-form on $X$ representing an integer cohomology class. In particular, integrating it against an integer homology class will give an integer invariant of $f$. – jgon Jun 22 '22 at 17:17
  • 1
    For 3, the answer should be no (at least if we stick to integrals that we get from de Rham cohomology stuff). For example, take $X=S^4$, $G=S^3$. Then $[X,G]=\mathbb{Z}/2$. However, for any $f:X\to G$, the induced maps on de Rham cohomology are 0 in dimension greater than 0, since the cohomology of $X$ in dimensions other than 0 and 4 is 0 and the cohomology of $G$ in dimensions other than 0 and 3 is 0. – jgon Jun 22 '22 at 17:28
  • @jgon Thank you so much for your dedicated answer.

    1. On your first comment, it is encouraging that we almost always have $G = SU(N)$ or $SO(N)$ in physics literature, so that $G$ is compact.

    2. Based on your answer, my question becomes more concrete as: "how to find a $k$-form representing an integer cohomology class?" Also, due to my lack of background, I want to ask what is the "integer homology class". Is it correct to say that $\int_N f^*\omega \in \mathbb Z$ whenever $N$ is a $k$-dimensional submanifold of $M$, in your settings?

    3. Thanks for a counterexample!

     – Laplacian Jun 23 '22 at 01:55
  • Another question on 3: Since you gave a counterexample, I want to ask for a specific case. In the URL "Why is the winding number of a matrix an integer?", a map $f: S^1 \to U(N)$ is classified by the degree of $\det \circ f: S^1 \to S^1$. Then, does this invariant completely characterizes $[S^1, U(N)]$? – Laplacian Jun 23 '22 at 01:57
  • By the way, I think your comment is very helpful so that it could be more appropriate as an answer. Also, could you suggest a reference to study these topics (how to find integer cohomology class, definition of integer homology class, etc)? – Laplacian Jun 23 '22 at 02:00

0 Answers0