# Why is dual lattice a lattice, in the context of complex tori

I have a simple linear algebra question regarding the definition of dual of a lattice; it was asked by someone else here three months ago on mathstackexchange but got no answer and few views, so forgive me for asking simple minded question here. Let $V$ be a complex vector space of dimension $n$, and let $L \subseteq V$ be a lattice (i.e a discrete free abelian subgroup) of rank $2n$. Let $$V^*=\{f:V \to \mathbb{C} \mid f(cv)=\overline{c}f(v) \}$$ be the space of $\mathbb{C}$-antilinear maps. This is a complex vector space of the same dimension as $V$ Let $$L^*=\{f \in V^* \mid Imaginary(f(L)) \subseteq \mathbb{Z} \}$$ be the elements of $V^*$ that map $L$ to complex numbers with integer imaginary part.
My question is why is $L^*$ a lattice in $V^*$? And why does it have the same rank as $L$? Obviously $L^*$ is a free abelian group. The issue is showing it is discrete in $V^*$. I tried choosing a $\mathbb{Z}$ basis for $L$ and writing out the condition for an to be in $L^*$, but it turned into a mess.

(The claims I am asking about are on p. 15 of Milne's notes on Abelian Varieties, here)

• It is obvious that $L^{\ast}$ is torsion-free, but freeness is not obvious until proving $L^{\ast}$ is discrete (forcing it to be finitely generated). Yet discreteness is obvious (so this is really not the issue): by $\mathbf{Z}$-finiteness of $L$ if such $f$ is "near 0" in $V^{\ast}$ then $f(L)$ has vanishing imaginary part, but $\mathbf{R}.L=V$ (!), so $f(V)=f(\mathbf{R}.L)=\mathbf{R}.f(L)\subset\mathbf{R}$, forcing the $\mathbf{C}$-linear $f$ to vanish. So the issue is really to show that the $\mathbf{R}$-span of $L^{\ast}$ coincides with $V^{\ast}$. Jan 3, 2016 at 4:14
• Hint for the $\mathbf{R}$-span aspect: $\mathbf{C} \otimes_{\mathbf{R}} V$ is $\mathbf{C}$-linearly naturally the direct sum of $V$ and its conjugate-space $\overline{V}$ (by decomposing $\mathbf{C} \otimes_{\mathbf{R}} \mathbf{C}$ as a $\mathbf{C}$-algebra for the left tensor-factor algebra structure), and $\mathbf{R} \otimes_{\mathbf{Z}} L = V$. Jan 3, 2016 at 4:15
• typo: I meant to say "conjugate-linear" near the end of my first comment (though the appearance of complex conjugation is a total red herring since composing such $f$'s with complex conjugation has no real effect on $L^{\ast}$ and so allows one to consider the original problem with $V^{\ast}$ taken to mean the usual $\mathbf{C}$-linear dual, thereby removing one layer of notational mess). Jan 3, 2016 at 4:24
• @nfdc23 nice argument to show discreteness, thanks! I don't understand your hint about how to show $\mathbb{R}$-span of $L^*$ is $V^*$. Jan 3, 2016 at 6:55
• I have written an extended answer below to explain what my hint was trying to get at. Jan 4, 2016 at 4:24

As explained in my comments, it is the same to treat the question where the notation $L^{\ast}$ is defined with $V^{\ast}$ taken to be the $\mathbf{C}$-dual (rather than the conjugate dual), so we do that. That is, $L^{\ast}$ now consists of the $\mathbf{C}$-linear (rather than conjugate-linear) forms on $V$ whose imaginary part is $\mathbf{Z}$-valued on $L$. Note also that $\mathbf{R} \otimes_{\mathbf{Z}} L$ coincides with the underlying $\mathbf{R}$-vector space $V_{\mathbf{R}}$ of $V$. We let $V_{\mathbf{R}}^{\ast}$ denote the $\mathbf{R}$-dual of $V_{\mathbf{R}}$ (not to be confused with the underlying $\mathbf{R}$-vector space of the $\mathbf{C}$-dual $V^{\ast}$!).
Write $\overline{V}$ to denote the conjugate-dual space (i.e., $\mathbf{C} \otimes_{\sigma, \mathbf{C}} V$ where $\sigma$ denotes complex conjugation), so for $v \in V$ we have the associated vector $\overline{v} := 1 \otimes v \in \overline{V}$. Thus, as $\mathbf{C}$-vector spaces we have naturally $$\mathbf{C} \otimes_{\mathbf{Z}} L = \mathbf{C} \otimes_{\mathbf{R}} V_{\mathbf{R}} \simeq V \oplus \overline{V}$$ where the isomorphism is defined by $1 \otimes v \mapsto (v, \overline{v})$. The inverse isomorphism is clearly $$(v, \overline{w}) \mapsto 1 \otimes \frac{v+w}{2} + i \otimes \frac{v-w}{2i}.$$ Now the idea is to $\mathbf{C}$-dualize both sides, exploiting that the $\mathbf{C}$-dual of $\mathbf{C} \otimes V_{\mathbf{R}}$ is naturally identified with $\mathbf{C} \otimes V^{\ast}_{\mathbf{R}}$, and that $V_{\mathbf{R}}^{\ast}$ is $\mathbf{R}$-spanned by the $\mathbf{Z}$-dual of $L$, with this $\mathbf{Z}$-dual identified as the $\mathbf{R}$-linear functionals on $V_{\mathbf{R}}$ that are $\mathbf{Z}$-valued on $L$. We'll thereby eventually find that up to a harmless factor of 2, this identifies the $\mathbf{Z}$-dual of $iL$ with $L^{\ast}$ as defined in the original problem (thereby "explaining" the notation) and from that we'll get the entire result (that $L^{\ast}$ as originally defined is both discrete in $V^{\ast}$ and spans it over $\mathbf{R}$) all at once.
To actually do the work, note that the restriction of the inverse isomorphism to the first direct summand is a $\mathbf{C}$-linear inclusion $V \hookrightarrow \mathbf{C} \otimes_{\mathbf{R}} V_{\mathbf{R}}$ given by $v \mapsto 1 \otimes (v/2) + i \otimes (v/2i)$. Thus, passing to $\mathbf{C}$-duals gives a $\mathbf{C}$-linear quotient map $\mathbf{C} \otimes_{\mathbf{R}} V_{\mathbf{R}}^{\ast} \twoheadrightarrow V^{\ast}$ defined by $$1 \otimes \lambda \mapsto (v \mapsto \lambda(v/2) + i \lambda(v/2i) = \lambda(v/2) - i \lambda(iv/2)).$$ The restriction to the $\mathbf{R}$-subspace $V_{\mathbf{R}}^{\ast}$ is an $\mathbf{R}$-linear map $V_{\mathbf{R}}^{\ast} \rightarrow V^{\ast}$ that is clearly injective and hence an isomorphism for $\mathbf{R}$-dimension reasons (or by bare hands: equivariance with $i$-scaling shows that $\mathbf{C}$-linear forms on $V$ can be uniquely written as $v \mapsto x(v) - ix(iv)$ for $\mathbf{R}$-linear forms $x$ on $V_{\mathbf{R}}$).
The above is really just natural context for something we could have written in a single line: $V_{\mathbf{R}}^{\ast} \simeq V^{\ast}$ via the formula $\lambda \mapsto (v \mapsto \lambda(v) - i\lambda(iv))$. Thus, for a lattice $L$ in $V$, the $\mathbf{Z}$-dual to $L$ as a lattice in the $\mathbf{R}$-dual $V_{\mathbf{R}}^{\ast}$ of $V_{\mathbf{R}} = \mathbf{R} \otimes_{\mathbf{Z}} L$ is carried over to the set of $\mathbf{C}$-linear forms $f$ on $V$ whose real part is $\mathbf{Z}$-valued on $L$, or equivalently (!!) whose imaginary part is $\mathbf{Z}$-valued on $iL$. In particular, we conclude that for any lattice $\Lambda$ in $V$, the set of $f \in V^{\ast}$ satisfying ${\rm{Im}}(f(i\Lambda)) \subset \mathbf{Z}$ is a lattice. Now take $\Lambda$ to be $iL$ (so $i\Lambda = L$) to conclude.