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Let $V$ be a vector space over the field of complex numbers, where scalar multiplication is written $\alpha {\cdot} v$. The complex conjugate space of $V$ is the vector space $\bar{V}$ which has the same underlying set and vector addition as $V$, but where multiplication by a scalar $\alpha$ is really multiplication by the complex conjugate of $\alpha$. Formally, the scalar multiplication operation $\bar{\cdot}$ of $\bar{V}$ is defined by: $$\alpha \bar{\cdot} v = \bar{\alpha} \cdot v$$

The map (on the category of complex vector spaces) which sends every space to its complex conjugate is a covariant functor: it sends a linear map $f : V \rightarrow W$ to itself, for the very same function $f$ is also a linear map $\bar{V} \rightarrow \bar{W}$.

Wikipedia claims that there is no natural isomorphism between $V$ and $\bar{V}$, but doesn't give a reference. Can anyone give me a hint for how this is rigorously proven?

Penelope Clairmont
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1 Answers1

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I think it is easier to see what is happening in a more general setting (but scroll down if you prefer a direct proof).

Let $f,g : A \to B$ be two ring homomorphisms. We have two forgetful functors $f^*, g^* : \mathbf{Mod}_B \to \mathbf{Mod}_A$ between the categories of left modules.

Claim. We have $f^* \cong g^*$ iff there is some $u \in B^{\times}$ such that $f = u \cdot g \cdot u^{-1}$ pointwise. In particular, if $B$ is commutative, $f^* \cong g^*$ holds iff $f = g$.

In our case, $f : \mathbb{C} \to \mathbb{C}$ is the identity and $g : \mathbb{C} \to \mathbb{C}$ is the complex conjugation. Notice that $f^*$ is the identity functor and $g^*$ is the "complex conjugate vector space"-functor. Since $f \neq g$, these functors are not isomorphic.

Proof of the claim. The functors $f^*,g^* : \mathbf{Mod}_B \to \mathbf{Mod}_A$ are isomorphic iff their left adjoint functors $\mathbf{Mod}_A \to \mathbf{Mod}_B$ are isomorphic. These are given by tensoring with the $(B,A)$-bimodules $(\mathrm{id}_B,B,f)$ and $(\mathrm{id}_B,B,g)$ (change of rings adjunction). By the (easy part of the) Eilenberg-Watts Theorem, the left adjoints are isomorphic iff these bimodules are isomorphic. So there is some bijective additive map $\alpha : B \to B$ such that $\alpha(bb') = b \alpha(b')$ (1) and $\alpha(b f(a)) = \alpha(b) g(a)$ (2) for all $b,b' \in B$ and $a \in A$. If $u := \alpha(1)$, then $\alpha(b) = b u$ by (1), and since $\alpha$ is bijective, $u \in B$ is a unit. Let's calculate both sides of (2). We have $\alpha(b f(a)) = b f(a) u$, and $\alpha(b) g(a) = b u g(a)$. So (2) is saying that $f(a) u = u g(a)$ for all $a \in A$, which means $f(a) = u g(a) u^{-1}$, and we are done.

See MO/37171 for a generalization to morphisms of schemes.

Direct proof. Let $\alpha : \mathrm{id} \to \mathrm{Conj}$ be any natural transformation from the identity functor to the "complex conjugate vector space"-functor on complex vector spaces. We will show $\alpha = 0$.

For every complex vector space $V$ then $\alpha_V : V \to V$ is a conjugate-linear map, meaning $\alpha_V(\lambda \cdot v) = \lambda^*\cdot \alpha_V(v)$, where $\lambda^*$ is the complex conjugate of $\lambda \in \mathbb{C}$.

Consider the linear map $f : V \to V$, $f(v) := \lambda \cdot v$. Naturality with respect to $f$ yields $$\alpha_V(\lambda \cdot v) = \alpha_V(f(v)) = f(\alpha_V(v))=\lambda \cdot \alpha_V(v).$$ So $\alpha_V$ is conjugate-linear and linear. Then $\lambda \cdot \alpha_V(v) = \lambda^* \cdot \alpha_V(v)$ for all $\lambda \in \mathbb{C}$. Apply this to any $\lambda \in \mathbb{C} \setminus \mathbb{R}$ to deduce $\alpha_V(v) = 0$. $\checkmark$

Martin Brandenburg
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