Disclaimer: Nothing in this note is original.

We will assume that the Yang-Mills potential $\omega_ {U}$ in patch $U$ has no singularity.

Let $\theta$ be the two-form curvature, we have the following observation for any vector bundle over all base manifolds,

\[\theta \wedge \theta = (d\omega+\omega \wedge \omega)\wedge (d\omega+\omega \wedge \omega) ,\]

using

\[\mathrm{Tr}\,(\omega \wedge \omega \wedge d\omega) = \mathrm{Tr}\,(d\omega \wedge \omega \wedge \omega)\]

and

\[\mathrm{Tr}\,(\omega \wedge \omega \wedge \omega \wedge \omega)=0\]

we have eventually

\[\mathrm{Tr}\,(\theta \wedge \theta) = d\, \mathrm{Tr}\,\left( \omega \wedge d\omega+\frac{2}{3}\omega \wedge \omega \wedge \omega \right)\]

thus $\mathrm{Tr}\,(\theta \wedge\theta)$ is locally exact, it is locally the differential of a $3$-form, called the Chern-Simons $3$-form. $\omega$ is not usually globally defined.

Coming back to the case of instanton, outside the instanton the curvature vanishes but the connection does not, hence we have

\[\int_ {U} \, \mathrm{Tr}\,(\theta \wedge \theta) = \int_ {\partial U = \mathbb{S}^{3}} \, \left( -\frac{1}{3} \right)\mathrm{Tr}\,(\omega \wedge \omega \wedge \omega).\]

This allows us to write the winding number in terms of curvature, by applying some kind of Stokes theorem

\[\frac{1}{24\pi^{2}}\int_ {\mathbb{S}^{3}} \, \mathrm{Tr}\,(\omega_ {U}\wedge \omega_ {U}\wedge \omega_ {U}) = - \frac{1}{8\pi^{2}} \int_ {\mathbb{R}^{4}} \, \mathrm{Tr}\,(\theta \wedge \theta).\]

Note that $\theta \wedge\theta$ is not the Lagrangian, which is $\mathrm{Tr}\,(\theta \wedge\star \theta)$. We have

\[\theta \wedge \theta = (\theta \wedge \theta)_ {0123}\,dt\wedge dx\wedge dy\wedge dz\propto \epsilon^{ijkl}F_ {ij}F_ {kl}\]

while

\[\theta \wedge \star \theta \propto F_ {ij}F^{ij}.\]

The Chern form for a $U(n)$ bundle

The topological significance of $\mathrm{Tr}\,(\theta \wedge\theta)$, generalizing Poincare’s theorem for closed surfaces (which says the sum of indices times multiplicity equal to the Euler characteristic), is developed by Chern. $\mathrm{Tr}\,(\theta \wedge\theta)$ is but one of a whole family of significant integrands, called the Chern forms. But before going into details, let’s review what is exterior power space shortly.

Given a vector space $V$ over a field $F$ (like the real numbers $\mathbb{R}$), the exterior algebra is an algebraic structure that extends the concept of scalars and vectors to higher-dimensional analogs. It is denoted as $\bigwedge V$. The $k$-th exterior power of $V$, denoted as $\Lambda^k(V)$ or $\bigwedge^{k} V$, is a vector space that consists of all alternating $k$-linear forms on $V$. For example, in $\Lambda^1(V)$, elements are just vectors. An exterior differential form of degree $k$ (or a $k$-form) on a differentiable manifold $M$ is a smooth section of the $k$-th exterior power of the cotangent bundle of $M$. In simpler terms, a $k$-form is a mathematical object that can be integrated over $k$-dimensional submanifolds of $M$. These forms are crucial in defining integrals over manifolds and in the formulation of Stokes’ theorem.

The exterior power space $\Lambda^k(V)$ provides the algebraic structure that underlies the concept of $k$-forms. When you consider a manifold $M$ with a tangent space at each point that is a vector space $V$, the exterior power $\Lambda^k(T^\ast M)$ (where $T^\star M$ is the cotangent bundle of $M$) is the space in which exterior differential forms live. This means that each $k$-form is an element of $\Lambda^k(T^\star M)$ at each point of $M$.

Example. 1

Considering the exterior powers of a three-dimensional complex vector space $V = \mathbb{C}^3$, with basis $\lbrace v_ {1},v_ {2},v_ {3} \rbrace$. We explore the spaces formed by taking the exterior powers of $V$. These spaces are constructed using the wedge product as follows:

Zeroth Exterior Power, $\Lambda^0(V)$: This is the space of scalars. It is isomorphic to the field over which the vector space is defined, in this case, the complex numbers $\mathbb{C}$. The dimension of this space is one.

First Exterior Power, $\Lambda^1(V)$: This is just the vector space $V$ itself, with dimension three (since $V = \mathbb{C}^3$). The basis are simply $\left\lbrace v_ {1},v_ {2},v_ {3} \right\rbrace$.

Second Exterior Power, $\Lambda^2(V)$: This space consists of all skew-symmetric bilinear forms on $V$. It can be visualized as the space of oriented planes in $V$. It has dimension

\[\binom{3}{2} = 3, \text{ since there are 3 ways to choose pairs from 3 elements.}\]

The basis are $\left\lbrace v_ {1}\wedge v_ {2},v_ {1}\wedge v_ {3},v_ {2}\wedge v_ {3} \right\rbrace$.

Third Exterior Power, $\Lambda^3(V)$: This is the space of 3-vectors in $V$. It represents the oriented volume elements in $V$. The dimension is $\binom{3}{3} = 1$ (there is only one way to choose triples from 3 elements). The basis is $v_ {1}\wedge v_ {2}\wedge v_ {3}$.

Higher Exterior Powers, $\Lambda^k(V)$ for $k > 3$: For any $k$ greater than the dimension of $V$ (which is 3 in this case), the exterior power $\Lambda^k(V)$ is the trivial vector space {0}, as there are no non-zero $k$-vectors in a 3-dimensional space for $k > 3$.

Let $V=\mathbb{C}^{N}$ be $N$-dimensional complex vector space. Let $A$ be a $N\times N$ complex matrix that acts on $V$ in the usual way. Consider the characteristic (eigenvalue) polynomial of $A$,

\[\det(\lambda I-A) = (\lambda-\lambda_ {1})(\lambda-\lambda_ {2})\dots(\lambda-\lambda_ {N})\]

where $\lambda_ {1},\dots,\lambda_ {N}$ are the eigenvalues of $A$. Putting $\lambda=-1$ we have

\[\begin{align*} \det(I+A) &= (1+\lambda_ {1})(1+\lambda_ {2})\dots(1+\lambda_ {N}) \\ &= 1 + \sum_ {i}\lambda_ {i} + \sum_ {i<j} \lambda _ {i} \lambda _ {j} +\dots+\prod_ {i=1}^{N}\lambda_ {i} \\ &= 1 + \mathrm{Tr}\,A + \mathrm{Tr}\, \bigwedge^{2}A +\dots+\mathrm{Tr}\,\bigwedge^{N}A \end{align*}\]

where the big wedges are called the elementary symmetric functions of he eigenvalues of $A$. The reason for this notation is as follows. Since

\[A: V \to V\]

is a linear transformation on $V$, we may let $A$ act on each of the exterior power spaces $\Lambda^{p} (A)$ by the so-called exterior power operation

\[\bigwedge^{p}A: \; v_ {1}\wedge \dots \wedge v_ {p} \mapsto (Av_ {1})\wedge (Av_ {2})\wedge \dots \wedge (Av_ {p} ).\]

Let’s take $p=N$ for example. $\Lambda^{N}(V)$ is one dimensional and

\[\left( \bigwedge^{N}A \right)(v_ {1}\wedge \dots \wedge v_ {n} ) = \det A\, (v_ {1}\wedge \dots \wedge v_ {n} ).\]

Thus justifies the definition shown in the last term in Eq. (1). Let’s take a look at other terms, for example, at $\Lambda^{2}A$. We treat $A$ as diagonal matrix $\text{diag}(\lambda_ {1},\dots,\lambda_ {N})$ to simplify the calculation, and we can consider

\[\bigwedge^{2}(v_ {i}\wedge v_ {j})\]

as the component of the exterior power operator on basis $v_ {i}\wedge v_ {j}$. Then the trace of $\Lambda^{2}(A)$ is simply the sum total of all its components. We have

\[\bigwedge^{2}(v_ {i}\wedge v_ {j}) = A_ {im} A_ {jn}(v_ {m} \wedge v_ {n} ) = (\lambda _ {i} \lambda _ {j})\, v_ {i} \wedge v_ {j}\]

where $i<j$ implicitly. Then the total sum is

\[\mathrm{Tr}\,\bigwedge^{2} A = \sum_ {i<j}\lambda _ {i} \lambda _ {j} .\]

It turns out we can further simplify $\Lambda^{p}A$. Note that

\[\lambda_ {1}^{p} + \lambda_ {2}^{p} + \dots + \lambda _ {N}^{p} = \mathrm{Tr}\,(A^{p}),\quad A = \text{diag}(\lambda_ {1},\dots,\lambda_ {N}).\]

Take $p=2$ for example,

\[\bigwedge^{2} A = \sum _ {i<j} \lambda _ {i} \lambda _ {j} = \frac{1}{2} \sum_ {ij} \lambda _ {i} \lambda _ {j} - \frac{1}{2} \lambda _ {i} \lambda _ {i} = \frac{1}{2}(\mathrm{Tr}\,A)^{2} - \frac{1}{2} \mathrm{Tr}\,A^{2}.\]

Turns out all the exterior power operators can be expressed as a polynomial in terms of $\mathrm{Tr}\, A^{n}$ and $(\mathrm{Tr}\,A)^{m}$ for some $m$ and $n$.

Now, let $E\to M$ be a complex $\mathbb{C}^{N}$ bundle with structure group $U(N)$. Let the connection be $\omega$. The corresponding $2$-form curvature is still denoted by $\theta$.

Let’s formally replace $A$ in Eq. (1) with $i\theta / 2\pi$, replace multiplication with wedge product. To be exact, we work with polynomial expression of $\Lambda^{p}A$, and perform these substitutions. Since $\theta$ is a $2$-form there will be no problem with ordering. Regarding $\theta$ as a matrix means that the $(\alpha,\beta)$-th entry of the matrix is $\theta^{\alpha}_ {\;\; \beta}$. Now similar to Eq. (1) we have

\[\begin{align*} \det \left( I+\frac{i\theta}{2\pi} \right) &= I + \mathrm{Tr}\,\frac{i\theta}{2\pi} + \dots \\ &= I + c_ {1}(E) + c_ {2}(E) + \dots \end{align*}\]

where $c_ {1}(E)$ is a 2-form on $U\subset M$, $c_ {r}(E)$ is a $2r$-form on $U$. It’s called the $r$-th Chern form.

To be specific, the form $c_ {1}$ is

\[c_ {1} = \frac{i}{2\pi} \mathrm{Tr}\,\theta = \frac{i}{2\pi} \mathrm{Tr}\,\theta^{\alpha}_ {\;\; \alpha}.\]

For $c_ {2}$ we have

\[c_ {2} = -\frac{1}{8\pi^{2}}[\mathrm{Tr}\,\theta \wedge \mathrm{Tr}\,\theta-\mathrm{Tr}\,(\theta \wedge \theta)].\]

Suppose the bundle has $SU(N)$ structure group rather than $U(N)$. The Lie algebra ${\frak su}(N)$ are traceless, anti-hermitian (mathematical convention) matrices, thus we have $\mathrm{Tr}\,\theta=0$ (recall that both the connection and curvature are ${\frak g}$-valued forms). Then $c_ {1}(E)$ vanishes but $c_ {2}$ does not,

\[c_ {2}(E) = \frac{1}{8\pi^{2}} \mathrm{Tr}\,(\theta \wedge \theta),\]

which is precisely the 4-form appearing in the winding number of an $SU(2)$ instanton! Just now we have $SU(N)$ instead.

Recall that the curvature $\theta_ {U}$ is locally defined on open patches $U$, under a change of basis the curvature changes as

\[\theta_ {V} = c_ {VU} \theta_ {U} c_ {UV}^{-1} ,\]

thanks to the property of determination we have

\[\det\left( I+ \frac{i \theta_ {V}}{2\pi} \right) = \det\left( I+ \frac{i\theta_ {U}}{2\pi} \right).\]

It means that each Chern form is globally define $2r$-form on all of $M^{n}$.

With the help of Bianchi identity, it is possible to show that $c_ {1}$ is closed.

For $SU(2)$ bundle, the second Chern form is locally the differential of Chern-Simons form.

We present without proof the

Theorem of Chern and Weil: Each $c_ {r}$ is a closed $2r$-form thus defines a (real) de Rham class. Furthermore, different connections for the $U(N)$ bundle will yield Chern form that differs by an exact form, and hence define the same de Rham cohomology class.