Because ${}^\star{}^\star=1$, the <exterior algebra> of two-forms has the orthogonal decomposition
$$
\Lambda^2=\Lambda^2_+\oplus\Lambda^2_-,
\qquad
F_\pm=\frac12(F\pm{}^\star F),
$$
where ${}^\star H=H$ for a <self-dual differential form> and ${}^\star G=-G$ for an <anti-self-dual differential form>. The Hodge star is self-adjoint, so
$$
\langle H,G\rangle
=\langle{}^\star H,{}^\star G\rangle
=-\langle H,G\rangle=0.
$$
It follows that
$$
\boxed{H\wedge G=-H\wedge{}^\star G
=-\langle H,G\rangle\operatorname{vol}=0}.
$$

For an $SU(n)$ connection, use the positive norm
$$
\|F\|^2=-\int_{\mathbb R^4}\operatorname{Tr}(F\wedge{}^\star F)
$$
and define the <Second Chern number> by
$$
k=-\frac1{8\pi^2}\int_{\mathbb R^4}\operatorname{Tr}(F\wedge F).
$$
Orthogonality gives
$$
\|F\|^2=\|F_+\|^2+\|F_-\|^2,
\qquad
8\pi^2k=\|F_+\|^2-\|F_-\|^2.
$$
Therefore the Euclidean <Yang-Mills action>
$$
S_{\rm YM}=\frac1{g_{\rm YM}^2}\|F\|^2
$$
obeys the <Yang-Mills instanton Bogomolny bound>
$$
\boxed{S_{\rm YM}\geq\frac{8\pi^2}{g_{\rm YM}^2}|k|}.
$$
Equality holds precisely when $F_-=0$ or $F_+=0$, according to the sign of $k$; these are the self-dual and anti-self-dual <Yang-Mills instantons>. Other trace and orientation conventions may reverse $k$ but leave the absolute-value bound unchanged.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 313 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-313/3/b/solution