Quaternionic representation

Question

Let $V$ be $G$-representation over quaternions $\mathbb{H}$. How to show that $$ \mathbb{H} \otimes_\mathbb{C} V $$ is canonically isomorphic to $V \oplus V$ as representation over $\mathbb{H}$? In tensor product $\mathbb{H}$ is viewed as right $\mathbb{C}$-module, and $V$ is viewed as left $\mathbb{C}[G]$-module.

Answer 1

Firstly, notice that representation $V$ over $\mathbb{H}$ is a complex representation and equivariant $\mathbb{R}$-linear map $J: V\to V,\,J^2=-1$, anticommuting with $I$, where $I$ is complex structure. If we replace $J$ by $-J$ we will get isomorphic representation with isomorphism given by $i$.

Let $V$ be quaternionic representation, $I$ and $J$ the corresponding maps. Let $W$ be $\mathbb{H} \otimes V$. On W there is well defined operator $$ f: W \to W, \, h \otimes v \mapsto hj\otimes Jv. $$ We should check only $$ f(hc\otimes v) = hcj\otimes Jv=hj\bar c\otimes Jv=hj\otimes \bar cJv=hj\otimes Jcv=f(h\otimes cv). $$ Obviously $f^2=1$ and $f$ is $\mathbb{H}$-linear. It easy to see that $W=W_1 \oplus W_2$, $$ W_1 = Im(f-1)=Ker(f+1), $$ $$ W_2 = Im(f+1) = Ker(f-1). $$ There are maps $$ a_1: V \to W_1, \, v\mapsto j\otimes Jv -1\otimes v, $$ $$ a_2: V \to W_2,\, v \mapsto j\otimes Jv + 1\otimes v. $$ They are complex-linear. Now look at the operator $J$: $$ a_1(Jv) = -j\otimes v-1\otimes Jv=ja_1(v) \Rightarrow a_1J=ja_1, $$ $$ a_2(Jv) = -j\otimes v + 1\otimes Jv = -ja_2(V) \Rightarrow a_2 J = -ja_2. $$ But we have seen at the beginning that the sign of $J$ does not matter.