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I am told in literature that if we have a continous map from a circle to unitary matrix

$$M : S^1 \to U(m)$$

then a winding number can be defined:

$$\nu=\frac{i}{2\pi}\int_0^{2\pi}dt\text{Tr}[M^{-1}(t)\partial_tM(t)]$$

Notice $M$ is a matrix. This number is an integer. Why would this number be an integer?

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  • $\begingroup$ after you integrate wont we get a matrix do you mean to say that the matrix has integer entries? $\endgroup$
    – happymath
    Commented Nov 21, 2016 at 10:53
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    $\begingroup$ @happymath Sorry, I missed a trace in the formula, see the edited post. $\endgroup$
    – atbug
    Commented Nov 21, 2016 at 11:09

1 Answer 1

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I am unfamiliar with your formula, but I understand why the answer should be an integer. Consider $$ det:U(n)\rightarrow S^1$$ the map given by the determinant. Since all the eigenvalues of a unitary matrix lie on the unit circle the map is well defined.

Given a loop in $\gamma:S^1\rightarrow U(n)$, the composition $det\circ \gamma:S^1\rightarrow S^1$ being a map from the circle to the circle has a degree. The degree is a signed count of the inverse image of a ``generic'' point on the circle and measures how many times $det\circ \gamma$ wraps around the circle. See the section of Munkres on the fundamental group of the circle.

Anyways, recall the principal branch of the logarithm is defined on the complement of the negative $x$-axis on the plane. Its restriction to the unit circle is $\mathbf{i}arg$, where $arg$ is the angular argument. Even though $arg$ is not globally defined on the circle its differential is. Note $dlog:TS^1\rightarrow \mathbf{i}\mathbb{R}$ is a purely imaginary one form whose integral over the circle is $2\pi{\mathbf{i}}$.

The pull back $det^*dlog$ to $U(n)$ is a one form. If you pull this back to the loop $\gamma:S^1\rightarrow S^1$ you get $$\gamma^*det^*dlog=(det\circ \gamma)^*dlog=d(log\circ det \circ \gamma)$$ is a one form on the circle. Change of variables says that $$\int_S^1(det\circ \gamma)^*dlog=deg(det\circ \gamma)2\pi{\mathbf{i}}.$$ Hence if you divide the integral by $2\pi\mathbf{i}$ you get an integer.

The formula you give comes from using the fact that if $M$ is an invertible matrix variable, $$d(log\circ det(M))=Tr(M^{-1}dM).$$ In fact, $$\frac{1}{2\pi{\bf i}}\int_\gamma Tr(M^{-1}dM)= \frac{1}{2\pi{\bf i}}\int_{S^1}d(log\circ det\circ \gamma)=deg(det\circ \gamma).$$

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  • $\begingroup$ What if the matrix $ M $ is a function as not only one variable, e.g., $ M(x, y) $ and $ M(x, y, z) $? $\endgroup$ Commented Dec 25, 2019 at 9:12
  • $\begingroup$ You need to add some conditions on the topology of the domain before you can have a meaningful answer. There are plenty of situations where there are. However it works better if you actually have a reason to do it. $\endgroup$ Commented Dec 26, 2019 at 10:07
  • $\begingroup$ Like Eq. (7) in this paper. $\endgroup$ Commented Dec 26, 2019 at 11:32
  • $\begingroup$ Indeed. The determinant of the matrices gives a map from your domain, call it $X$ to the circle $S^1$. Homotopy classes of maps from $X$ to $S^1$, describe $H^1(X;\mathbb{Z})$, where you are integrating the pullback of the form in the paper along a cycle in your space $X$ to evaluate it. The answer will always be an integer. $\endgroup$ Commented Dec 26, 2019 at 12:22

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