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Does there exists an $\mathbb R$-linear embedding $\bigwedge^k_{\mathbb{C}}\mathbb{C}^d \to \bigwedge^k_{\mathbb{R}}\mathbb{C}^d$ which maps decomposable tensors to decomposable tensors?

(The subscripts specify over which field we are taking the tensor/exterior powers. )

I specifically want an embedding which respects the "Grassmannian" structures of the exterior powers. Some embedding always exists, due to dimensional reasons:

$\dim_{\mathbb R}(\bigwedge^k_{\mathbb{C}}\mathbb{C}^d)=2\binom{d}{k} \le \binom{2d}{k} =\dim_{\mathbb R}(\bigwedge^k_{\mathbb{R}}\mathbb{C}^d).$

The naive candidate would be to take $v_1 \wedge \dots \wedge v_k$ to "itself". However, this does not really define a well-defined map $\bigwedge^k_{\mathbb{C}}\mathbb{C}^d \to \bigwedge^k_{\mathbb{R}}\mathbb{C}^d$. For $k=2$, $iv_1 \wedge v_2=v_1 \wedge iv_2$ are equal as elements of $\bigwedge^k_{\mathbb{C}}\mathbb{C}^d$, but not are not always equal as elements of $\bigwedge^k_{\mathbb{R}}\mathbb{C}^d$:

Set $v_1=\begin{equation} \left(\matrix{ 1 \cr i\cr }\right) \end{equation},$ $ v_2=\begin{equation} \left(\matrix{ -i \cr 0\cr }\right) \end{equation}$. Then $\text{span}_{\mathbb R}(v_1,iv_2)=\text{span}_{\mathbb R}(\left(\matrix{ 1 \cr i\cr }\right),\left(\matrix{ 1 \cr 0\cr }\right)) \neq \text{span}_{\mathbb R}(\left(\matrix{ i \cr -1\cr }\right),\left(\matrix{ -i \cr 0\cr }\right)) = \text{span}_{\mathbb R}(iv_1,v_2)$, so $v_1 \wedge iv_2 , iv_1 \wedge v_2$ are not equal as elements of $\bigwedge^k_{\mathbb{R}}\mathbb{C}^d$.

Comment: I don't really know how to construct $\mathbb{R}$-linear maps from $\bigwedge^k_{\mathbb{C}}\mathbb{C}^d$, since the usual universal property of exterior power, only give a way to construct $\mathbb{C}$-linear maps.

Asaf Shachar
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  • Note that $\bigwedge_{\Bbb{C}}^k\Bbb{C}^d$ is naturally a quotient of $\bigwedge_{\Bbb{R}}^k\Bbb{C}^d$. Are you looking for a section of the quotient map? – Servaes Feb 04 '19 at 13:44
  • Well, I am looking for an $\mathbb R$-linear embedding which maps decomposable tensors to decomposable tensors, and takes $\bigwedge^k {\mathbb R}^d$ inside $\bigwedge^k_{\mathbb{C}}\mathbb{C}^d $ to its counterpart in $\bigwedge^k_{\mathbb{R}}\mathbb{C}^d$. (You can see here: https://math.stackexchange.com/questions/3098324/if-omega-in-bigwedgek-mathbbrd-is-decomposable-over-mathbb-c-is-it... for further motivation and explanation of how I view $\bigwedge^k {\mathbb R}^d$ as a subspace of both $\bigwedge^k_{\mathbb{C}}\mathbb{C}^d $ and $\bigwedge^k_{\mathbb{R}}\mathbb{C}^d$). – Asaf Shachar Feb 04 '19 at 13:48
  • By the way, can you elaborate on how $\bigwedge_{\Bbb{C}}^k\Bbb{C}^d$ is realized as a quotient of $\bigwedge_{\Bbb{R}}^k\Bbb{C}^d $? I don't see it... – Asaf Shachar Feb 04 '19 at 14:43

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