# Conjugation in the complexification of a vector space switches its type

Let $V$ be a real vector space with an almost complex structure $J$ and consider its complexification $V^\mathbb{C}$ where we extend $\mathbb{C}$-linearly the linear maps of $V$, in particular $J$. In this space we define $V^{1,0}$ as the $+i$ eigenspace of $J^\mathbb{C}$ (the extension of $J$) and $V^{0,1}$ as the $-i$ eigenspace.

I want to show that the complex conjugation of a vector in $V^{1,0}$ gives me a vector in $V^{0,1}$. Remember that the conjugation of $V^\mathbb{C} \ni v = w \otimes \lambda$ (where $w \in V$), is $w \otimes \overline{\lambda}$.

Let $v$ be in $V^{1,0}$, then $J^\mathbb{C} v = i v$, that is, $J^\mathbb{C}(w\otimes \lambda )= J^\mathbb{C}(w\otimes 1) \lambda = i (w\otimes 1) \lambda$ so \begin{equation}J^\mathbb{C}(w\otimes 1) = w\otimes i \end{equation} now $J^\mathbb{C} \overline{v} = J^\mathbb{C} (w \otimes \overline{\lambda} ) = J^\mathbb{C} (w \otimes 1)\overline{\lambda}$ using the last equation we obtain, \begin{equation} J^\mathbb{C} \overline{v} = w\otimes i \overline{\lambda} =i \overline{v} \end{equation} Where am I missing a sign? Thanks.

Your mistake was to assume $v \in V^{1,0}$ is of the form $w\otimes\lambda$.

Recall that $V^{\mathbb{C}} = V\otimes_{\mathbb{R}}\mathbb{C}$ consists of finite sums of elements of the form $w\otimes\lambda$. As $\mathbb{C}$ is a two-dimensional vector space over $\mathbb{R}$ with basis $\{1, i\}$, every element of $V^{\mathbb{C}}$ (and hence $V^{1,0}$) is of the form $w_1\otimes 1 + w_2\otimes i$ for some $w_1, w_2 \in V$.

The vector $v = w_1\otimes 1 + w_2\otimes i \in V^{\mathbb{C}}$ belongs to $V^{1,0}$ if and only if $J^{\mathbb{C}}(v) = iv$. As

$$J^{\mathbb{C}}(v) = J^{\mathbb{C}}(w_1\otimes 1 + w_2\otimes i) = J(w_1)\otimes 1 + J(w_2)\otimes i$$

and

$$iv = i(w_1\otimes 1 + w_2\otimes i) = w_1\otimes i - w_2\otimes 1,$$

we see that $v \in V^{1,0}$ if and only if $J(w_1) = -w_2$ and $J(w_2) = w_1$; note, each equality implies the other as $J^2 = -\operatorname{id}_V$. Therefore, we see that $V^{1,0} = \{w\otimes 1 - J(w)\otimes i \mid w \in V\}$.

One can do as above for $V^{0,1}$ and find that $V^{0,1} = \{w\otimes 1 + J(w)\otimes i \mid w \in V\}$. Now one sees that given $v \in V^{1,0}$ we have

$$\bar{v} = \overline{w\otimes 1 - J(w)\otimes i} = w\otimes\bar{1} - J(w)\otimes\bar{i} = w\otimes 1 + J(w)\otimes i \in V^{0,1}.$$

The difference between $J$ and $J^C$ lies in the fact that the tensor product does not let $i$ flow from one side to the other. The operation $J$ is multiplying by $i$ on the left of the tensor, which changes the positions of the elements of $w$ in the basis $\{e_i, ie_i\}$ and signs of half of them: $i.\Sigma[ a_i e_i + b_i(i e_i)] = \Sigma [-b_i e_i + a_i (ie_i)]$. The operation $J^C$ is multiplying by $i$ on the right of the tensor, which swaps the positions of $w_1$ and $w_2$ while negating one of them. This difference between $J$ and $J^C$ leads to the minus sign.