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.