Appendix A

Proof. For \(v=0\) the statement is obvious so let’s assume \(v \neq 0\). Let \(w_1,w_2\in W\) be linearly independent, write \(g(v\otimes w_1)=v_1\otimes u_1\) and \(g(v\otimes w_2)=v_2\otimes u_2\) for some vectors \(v_1,v_2 \in V\) and \(u_1,u_2 \in W\) all nonzero. Then \(g \in G(m,n)\) implies that one of the two following cases occurs:

• \(v_1\wedge v_2=0\) and \(u_1\wedge u_2\neq 0,\)

• \(v_1\wedge v_2\neq 0\) and \(u_1\wedge u_2=0\).

Assume (I) holds and fix \(v_0\neq 0\) with \(v_0\wedge v_1=0\). It follows again with \(g \in G(m,n)\), that for every \(w \in W\), there exists a unique element \(b_v(w)\in W\) such that \(g(v\otimes w)=v_0\otimes b_v(w)\). The assignment \(w \mapsto b_v(w)\) is invertible and linear, thus (i) follows. Assuming (II) holds we conclude similarly that (ii) follows.

Suppose case (i) occurs for some \(v \in V\) and case (ii) for some \(v^{\prime} \in V\). Let \(b_v\) and \(a_{v^{\prime}}\) be the associated isomorphisms. Since \(g \in G(m,n)\) either \(b_v\) or \(a_{v^{\prime}}\) must have rank \(1\), thus contradicting the fact that both maps are isomorphisms.

Proof. Let \(g \in G(m,n)\). Assume case (i) of Lemma A.1 holds for some and hence all \(v \in V\). Let \(\hat v,\tilde{v} \in V\), then by Lemma A.1 there exists \(\hat v_0,\tilde{v}_0 \in V\) and \(b_{\hat{v}},b_{\tilde{v}} \in \mathrm{Isom}(W,W)\) such that for all \(w \in W\) we have \[g(\hat{v}\otimes w)=\hat{v}_0\otimes b_{\hat{v}}(w), \quad \varphi(\tilde{v}\otimes w)=\tilde{v}_0\otimes b_{\tilde{v}}(w).\] On the other hand for some \(\tilde{w} \in W\) there must exist \(\tilde{w}_0 \in W\) and \(a_{\tilde{w}} \in \mathrm{Isom}(V,V)\) such that for all \(v \in V\) \[g(v\otimes \tilde{w})=a_{\tilde{w}}(v)\otimes \tilde{w}_0.\] We thus get \[g(\hat{v}\otimes \tilde{w})=a_{\tilde{w}}(\hat{v})\otimes \tilde{w}_0=\hat{v}_0\otimes b_{\hat{v}}(\tilde{w})\] and \[g(\tilde{v}\otimes \tilde{w})=a_{\tilde{w}}(\tilde{v})\otimes \tilde{w}_0=\tilde{v}_0\otimes b_{\tilde{v}}(\tilde{w})\] Since this holds for any \(\tilde{w} \in W\), the map \(b_{\tilde{v}}\) must be a (nonzero) constant multiple of the map \(b_{\hat{v}}\). It follows that there exists \(a \in \mathrm{Isom}(V,V)\) and \(b \in \mathrm{Isom}(W,W)\) such that \(g(v\otimes w)=a(v)\otimes b(w)\) for all \(v\in V\) and \(w \in W\). If the case (ii) of Lemma A.1 holds, we can conclude similarly that there exists \(a \in \mathrm{Isom}(W,V)\) and \(b \in \mathrm{Isom}(V,W)\) and such that \(g(v\otimes w)=a(w)\otimes b(v)\) for all \(v \in V\) and \(w \in W\). From this the claim follows easily.