# 12-10 Notes

In this final lecture, we wrap up our discussion of the construction of $k$-instantons over $S^4$.

Let $C,D$ two be $(n+k) \times k$ matrices. Consider the matrix $Cx+Dy$, with $x,y \in \mathbb{H}$. Its columns determine for us a quaternionic k-plane in $\mathbb{H}^{n+k}$. We get a bundle $E$ over $\mathbb{PH}^1 = S^4$, where the fiber $E|_{[x:y]}$ is the orthogonal complement of $Cx + Dy$ in $H^{n+k}$. We get a connection on $E$ induced from the product connection on $\mathbb{H^{n+k}} \times S^4$.

Write $v(x,y) = Cx+Dy$. Then $Q = v(v^*v)^{-1}v^*$ is the orthogonal projection onto the span of $Cx+Dy$. Set $y = 1$. Let $P = I - Q$ be the complementary projection. The curvature our connection can be written as $$F = PC dx(Cx+D)^*(Cx+D)^{-1}\wedge d\bar x C^*)P.$$ Since $dx \wedge d\bar x$ is self-dual, then $F$ is self-dual if and only if $$(Cx+D)^*(Cx+D)^{-1} \textrm{\;is\;real.}$$

More explicitly, let $C = \begin{pmatrix}C_0 \\ C_1\end{pmatrix}$, $D = \begin{pmatrix}D_0 \\ D_1\end{pmatrix}$, where $C_0,D_0$ are $n \times k$ and $C_1,D_1$ are $k\times k$. Observe that the above reality is condition is preserved under left of multiplication of $Cx + D$ by an element of $Sp(n+k)$, or right multiplication by an element of $O(k)$. One can therefore check that without loss of generality, we can take $C_0 = O, C_1 = -I.$ write $D = \begin{pmatrix}\Lambda \\ B\end{pmatrix}$. Then the expression in (1) becomes $\Lambda^*\Lambda + B^*B - (\bar x B + B^* x) + |x|^2I$. Thus, for this expression to be real for all $x$, we need (i) $\Lambda^*\Lambda + B^*B$ to be real (ii) $\bar x B + B^*x$ is real $\iff B^T = B$.

Suppose $n=1$, $\Lambda = [\lambda_1, \ldots, \lambda_n]$. If we do a formal dimension count, then $\dim \Lambda = 4k$, $\dim B = 4k(k+1)/2 = 2k^2 + 2k$, condition (i) imposes $3k(k-1)/2$ constraints, and the dimension of the symmetry group $O(k) \times Sp(1)$ is $k(k-1)/2 + 3$. So overall, the formal dimension of the moduli space of $SU(2)$ $k$-instantons is $$d_k = (4k + 2k^2+2k) - (3k(k-1)/2 + k(k-1)/2 + 3) = 8k-3.$$ This is the correct answer in fact as can be shown rigorously using the Atiyah-Singer index theorem.