A question in spin geometry in dimension 8

Question

$\DeclareMathOperator\trace{trace}\DeclareMathOperator\End{End}\DeclareMathOperator\Trace{Trace}$This is to understand a very specific isomorphism in dimension $8$. In dimension $4$ for a spin$^c$ manifold $M^4$ and $S=S_+\oplus S_-\rightarrow M$ a spinor bundle, we have the following isomorphism of normed vector spaces given by the Clifford multiplication: \begin{align*} c:i\Omega^2_+\rightarrow i\mathfrak{su}(\Gamma(S_+)) \end{align*} the norm on the rhs being \begin{align*} \langle T,S\rangle:=\frac{1}{4}\text{trace}(TS^*). \end{align*} I wish to understand the similar isomorphism in dimension $8$. In dimension $8$ the following is an isomorphism of two normed vector spaces: \begin{align*} c:i\Omega^2\oplus\Omega^4_+\rightarrow i\mathfrak{su}(\Gamma(S_+)). \end{align*} $c$ denotes Clifford multiplication. The norm on the lhs is the usual norm on forms induced by the Riemannian metric and the norm on the rhs is the norm from $\End(\Gamma(S_+))$, i.e., for $T,S\in \End(\Gamma(S_+))$, \begin{align*} \langle T,S\rangle:= a\trace(TS^*). \end{align*} For $T\in i\mathfrak{su}(\Gamma(S_+))$, $T=T^*$. Hence, \begin{align*} |T|^2=a\trace(T^2). \end{align*} We need to figure out what is $a$. Let's check it in the Kähler case. $S_+=\Omega^0\oplus\Omega^{0,2}\oplus\Omega^{0,4}$, \begin{align*} c(i\omega)=\begin{bmatrix} 4&0&0\\ 0&0&0\\ 0&0&-4 \end{bmatrix}, c(\omega^2)=\begin{bmatrix} -12&0&0\\ 0&4&0\\ 0&0&-12 \end{bmatrix} \end{align*} \begin{align*} c(i\omega)^2=\begin{bmatrix} 16&0&0\\ 0&0&0\\ 0&0&16 \end{bmatrix}, c(\omega^2)^2=\begin{bmatrix} 144&0&0\\ 0&16&0\\ 0&0&144 \end{bmatrix}\\ \frac{1}{8}\Trace\bigl(c(i\omega)^2\bigr)=\frac{32}{8}=4=|\omega|^2\\ \frac{1}{16}\Trace\bigl(c(\omega^2)^2\bigr)=\frac{144\times 2+6\times 16}{16}=24=|\omega^2|^2. \end{align*} It seems $a=\frac{1}{8}$ for $T\in c(i\Omega^2)$ and $a=\frac{1}{16}$ for $T\in c(\Omega^4_+)$. Now it seems a bit odd as I was hoping to get the same constant. Given $\phi\in\Gamma(S_+)$, $q(\phi):=\phi\otimes\phi^*-\frac{1}{8}|\phi|^2\mathrm{Id}\in i\mathfrak{su}(\Gamma(S_+))$. How do we calculate the norm of this element?

Answer 1

I have an answer when the structure is Spin and not just Spin$^c$. The constants $\frac{1}{8}$ and $\frac{1}{16}$ are indeed correct, one can check in orthonormal frame as well.

For $\theta\in i\Omega^2,$ we get \begin{align*} &\langle\theta,q(\phi)\rangle\\ &=\frac{1}{8}\langle c(\theta),\phi\otimes\phi^*-\frac{1}{8}|\phi|^2\rangle\\ &=\frac{1}{8}\text{trace}\big(c(\theta)\circ(\phi\otimes\phi^*-\frac{1}{8}|\phi|^2)\big)\\ &=\frac{1}{8}\text{trace}\big(c(\theta)\circ (\phi\otimes\phi^*)\big)\hspace{1 ex}[\text{since }c(\theta) \text{ is trace-free}]\\ &=\frac{1}{8}\sum_i\langle c(\theta)(\langle e_i,\phi\rangle\phi),e_i\rangle \hspace{1 ex}[e_i\text{s form an orthonormal basis of $S_+$ point wise}]\\ &=\frac{1}{8}\sum_i\langle c(\theta)\phi,(\langle\phi,e_i\rangle e_i)\rangle\\ &=\frac{1}{8}\langle c(\theta)\phi,\phi\rangle \end{align*} and similarly for $\theta\in\Omega^4_+,$ we get \begin{align*} \langle \theta,q(\phi)\rangle=\frac{1}{16}\langle c(\theta)\phi,\phi\rangle \end{align*} $S_+$ has a Spin$(8)$-equivariant real structure. We write $S_+=S_\mathbb{R}\oplus iS_\mathbb{R},\phi=\phi_1\oplus i\phi_2,\phi_j\in\Gamma(S_\mathbb{R}).$ \begin{align*} &\phi^*\phi-\frac{1}{8}|\phi|^2\\ &=(\phi_1+i\phi_2)^*(\phi_1+i\phi_2)-\frac{1}{8}(|\phi_1|^2+|\phi_2|^2)\\ &=\big(\phi_1^*\phi_1+ \phi_2^*\phi_2-\frac{1}{8}(|\phi_1|^2+|\phi_2|^2)\big)+i(\phi_1^*\phi_2-\phi_2^*\phi_1) \end{align*} $c(i\Omega^2)$ anti-commutes with this real-structure and $c(\Omega^4_+)$ commutes with this real-structure. Hence, \begin{align*} c^{-1}\big(\phi_1^*\phi_1+ \phi_2^*\phi_2-\frac{1}{8}(|\phi_1|^2+|\phi_2|^2)\big)\in\Omega^4_+\\ c^{-1}\big(i(\phi_1^*\phi_2-\phi_2^*\phi_1)\big)\in i\Omega^2 \end{align*} \begin{align*} &||q(\phi)||^2\\ &=\langle q(\phi)_2,q(\phi)\rangle+\langle q(\phi)_4,q(\phi)\rangle\\ &=\frac{1}{8}\langle c(q(\phi)_2)\phi,\phi\rangle+\frac{1}{16}\langle c(q(\phi)_4)\phi,\phi\rangle \end{align*} \begin{align*} &\langle q(\phi)_4\phi,\phi\rangle\\ &=\langle |\phi|^1\phi_1+|\phi_2|^2i\phi_2+\langle\phi_1,\phi_2\rangle(\phi_2+i\phi_1)-\frac{1}{8}(|\phi_1|^2+|\phi_2|^2)(\phi_1+i\phi_2),(\phi_1+i\phi_2)\rangle\\ &=|\phi_1|^4+|\phi_2|^4+2|\langle\phi_1,\phi_2\rangle|^2-\frac{1}{8}(|\phi_1|^2+|\phi_2|^2)^2 \end{align*} \begin{align*} &\langle q(\phi)_2\phi,\phi\rangle\\ &=\langle |\phi_1|^2i\phi_2-\langle\phi_1,\phi_2\rangle i\phi_1-\langle\phi_1,\phi_2\rangle\phi_2+|\phi_2|^2\phi_1,\phi_1+i\phi_2\rangle\\ &=2\big(|\phi_1|^2|\phi_2|^2-|\langle\phi_1,\phi_2\rangle|^2\big) \end{align*} Thereafter, \begin{align*} &||q(\phi)||^2\\ &=\frac{1}{8}\langle c(q(\phi)_2)\phi,\phi\rangle+\frac{1}{16}\langle c(q(\phi)_4)\phi,\phi\rangle\\ &=\frac{1}{4}\langle \big(|\phi_1|^2|\phi_2|^2-|\langle\phi_1,\phi_2\rangle|^2\big)+\frac{1}{16}\big(|\phi_1|^4+|\phi_2|^4+2|\langle\phi_1,\phi_2\rangle|^2-\frac{1}{8}(|\phi_1|^2+|\phi_2|^2)^2\big)\\ &=\frac{1}{16}\Big(\frac{7}{8}(|\phi_1|^4+|\phi_2|^4)+\frac{15}{4}|\phi_1|^2|\phi_2|^2-2|\langle\phi_1,\phi_2\rangle|^2\Big)\\ &=\frac{1}{8\times 16}\big(7(|\phi_1|^4+|\phi_2|^4)+30|\phi_1|^2|\phi_2|^2-16|\langle\phi_1,\phi_2\rangle|^2\big)\\ &=\frac{1}{8\times 16}\big(7(|\phi_1|^2+|\phi_2|^2)^2+16(|\phi_1|^2|\phi_2|^2-|\langle\phi_1,\phi_2\rangle|^2)\big) \end{align*}