Calculation of the Pontryagin index

[S_klogW]

When I read A.Zee's QFT in a Nutshell, he asked me to do the calculation of his exercise IV.4.6: (A.Zee called these the Pontryagin Index)

Let g(x) be the element of a group G. The 1-form v = gdg† is known as the Cartan-Maurer form. Then tr v^N is trivially closed on an N-dimensional manifold since it is already an N-form. Consider Q = SN ∫tr vN with SN the N-dimensional sphere. Discuss the topological meaning of Q. These con- siderations will become important later when we discuss topology in field theory in chapter V.7. [Hint: Study the case N = 3 and G = SU(2).]

I found that the result for G=SU(2) and N=3 is -24π^2, is it correct? I also calculated for the case N=1 and G=U(1), and i found the result is -2πi. What's on Earth is the topological meaning of these results?

Thanks.

 

[Bill_K]

Isn't v^N the volume element? Sounds like you're calculating the group volume.

 

[tom.stoer]

Just to clarify the notation: you have to calculate

$$v_i = g^\dagger \partial_{i}\,g$$

$$Q[g] = \frac{1}{c}\int_{S^N}d^N\Omega\,\epsilon^{i_1 i_2 \ldots i_N}\,\text{tr}(v_{i_1}\,v_{i_2}\ldots v_{i_N})$$

for various Lie groups G with a Lie group valued function

$$g \in G$$

I guess the main problem you have is to find the correct constant c which obviously depends on the group G.

 

[tom.stoer]

OK, let's continue with U(1) on S¹. Let's write

$$g(x) = e^{i\nu(x)}$$

with a function nu(x) respecting the S¹ periodicity, i.e. with

$$\nu(x+2\pi) = \nu(x)+n\;\Rightarrow\;g(x+2\pi) = g(x)+n$$

Calculating v(x) we get

$$v_x = g^\dagger\,\partial_x\,g = i\,\partial_x\,\nu$$

For the integral we find

$$-i\int_0^{2\pi}dx \, v_x = \int_0^{2\pi}dx \, \partial_x\,\nu = \nu(2\pi) - \nu(0) = n$$

That means that the function nu(x) runs from 0 to n when x runs from 0 to 2π; so g runs around the U(1) circle n-times when x runs around the S¹ circle once. That means that

$$Q[g] = n$$

is nothing else but the winding number of the map

$$g: S^1 \to U(1)$$