# Is there a natural way to view $\bigwedge^k_{\mathbb{C}}\mathbb{C}^d$ as a subspace of $\bigwedge^k_{\mathbb{R}}\mathbb{C}^d$

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 $$v_1=$$\left(\matrix{ 1 \cr i\cr }\right)$$,$$ $$v_2=$$\left(\matrix{ -i \cr 0\cr }\right)$$$$. Then $$\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.

• 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 4 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: math.stackexchange.com/questions/3098324/…... 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 4 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 4 at 14:43