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For context, this question comes from my reading of Kobayashi+Nomizu's differential geometry book, volume 2 (pages 116-117).

Given a real vector space $V$ with $\mathrm{dim}(V)=2n$, a complex structure is a linear endomorphism satisfying $J^2=-1$.

One can also define the complexification of a vector space as $V^{\mathbb{C}} = V \otimes_{\mathbb{R}} \mathbb{C}$.

Now, $J$ extends naturally to a complex endomorphism on the complexified space, and (Kobayashi and Nomizu claim that) it has eigenvalues $\pm i$.

I have some confusions about this:

  1. Is it true that the natural extension of $J$ is just defined by $J(V\otimes z) = J(V)\otimes z$?
  2. More crucially, the complexified vector space is treated as a tensor product over the reals, so isn't the coefficient field of this new space also the reals? Isn't, then, the characteristic polynomial viewed as a polynomial over $\mathbb{R}$ (and thus cannot have imaginary roots)?
  3. My intuition is to treat application of $J$ as multiplication by $i$ in the original space, but the complexified space offers a bona fide multiplication by $i$, and I'm not sure how these two interact. What would be an example of a $v\in V^{\mathbb{C}}$ such that $Jv = -iv$? What does $Jv=iv$ actually mean? I've tried to answer these questions for $V=\mathbb{R}^2$ with the canonical structure but I'm coming up empty handed.

I greatly appreciate your time and help!

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    $\begingroup$ Are you claiming that a polynomial with real coefficients cannot have imaginary roots? $\endgroup$
    – José Carlos Santos
    Commented Aug 29, 2020 at 15:32
  • $\begingroup$ Here is a broader context. Suppose $V$ is a vector space over a field $K$, $A \colon V \to V$ is a $K$-linear map, and $L/K$ is an extension field. Do you know how to make the $K$-vector space $V \otimes_K L$ into an $L$-vector space of the same $L$-dimension and the $K$-dimension of $V$? Do you know why $A$ extends to a $K$-linear map $V \otimes_K L \to V \otimes_K L$ where $A(v \otimes c) = Av \otimes c$ on simple tensors and why this is actually $L$-linear using the $L$-vector space structure on $V \otimes_K L$? $\endgroup$
    – KCd
    Commented Aug 29, 2020 at 15:32
  • $\begingroup$ @KCd I believe so. Scalar multiplication on simple tensors is $l_1(v\otimes l_2) = v \otimes l_1 l_2$, and you can extend this to any element by the distributive property. The $L$ dimension is the same as the $K$ dimension of $V$ because if $\{v_i\}$ is a basis for $V$, then $\{v_i \otimes 1\}$ is an $L$ basis for $V\otimes_K L$. The extension map is $L$ linear because $A( l_1 v) = A(v\otimes l_1) = Av\otimes l_1 = l_1(Av\otimes 1) = l_1 Av$ $\endgroup$
    – npdotrand
    Commented Aug 29, 2020 at 16:02
  • $\begingroup$ It is not strictly correct that a mapping on a tensor product space is "extended to any element by the distributive property" from a formula on simple tensors: if there is a linear map then its values on simple tensors determine its values everywhere, but merely declaring values on simple tensors is not a way to define anything because a tensor can be written as a sum of simple tensors in lots of ways. The way to define the mapping on a tensor product space is by the universal property (unless you're a physicist, I guess). $\endgroup$
    – KCd
    Commented Aug 29, 2020 at 16:16
  • $\begingroup$ For example, you can't define a $K$-linear mapping $V \otimes_K V \to V$ by $f(v_1 \otimes v_2) = v_1 + v_2$ and then "extend to any element by the distributive property". $\endgroup$
    – KCd
    Commented Aug 29, 2020 at 16:19

1 Answer 1

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After @KCd wrote the helpful comments regarding the broader context of extending vector spaces, I believe I've come up with the answers, so for completeness to this question I'll write them up here:

  1. Yes.

  2. This seems to be true on a technicality, but because $V^{\mathbb{C}}$ can also be viewed as a $\mathrm{dim}_{\mathbb{R}}(V)$ space over the complex numbers, we can think of the characteristic polynomial over $\mathbb{C}$, so $\lambda= \pm i$ is perfectly fine.

  3. Once I stopped tunnel visioning on simple tensors, the breakdown became clear. For the example of $V=\mathbb{R}^2$ with the canonical complex structure, then in $V^{\mathbb{C}}$, $$J\left(\begin{pmatrix} 1 \\\ 0 \end{pmatrix} \otimes 1 - \begin{pmatrix} 0 \\\ 1 \end{pmatrix} \otimes i \right) = \begin{pmatrix} 0 \\\ 1 \end{pmatrix} \otimes 1 + \begin{pmatrix} 1 \\\ 0 \end{pmatrix} \otimes i = i\left ( \begin{pmatrix} 1 \\\ 0 \end{pmatrix} \otimes 1 - \begin{pmatrix} 0 \\\ 1 \end{pmatrix} \otimes i \right )$$ exhibits an element of the positive eigenspace $V^{1,0}$. Of course, the notation simplifies tremendously if you think of the space as $V(\mathbb{C})$ rather than $V(\mathbb{R})\otimes_{\mathbb{R}} \mathbb{C}$, but computing with the explicit tensor product was quite beneficial to my understanding.

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