# Eigenvalues of Complex Stucture in a Complexified Vector Space

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 $$\pm i$$.

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!

• Are you claiming that a polynomial with real coefficients cannot have imaginary roots? Commented Aug 29, 2020 at 15:32
• 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 $V \otimes_K L$?
– KCd
Commented Aug 29, 2020 at 15:32
• @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 $A( l_1 v) = A(v\otimes l_1) = Av\otimes l_1 = l_1(Av\otimes 1) = l_1 Av$ Commented Aug 29, 2020 at 16:02
• 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).
– KCd
Commented Aug 29, 2020 at 16:16
• 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".
– KCd
Commented Aug 29, 2020 at 16:19

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.