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)

  • 3
    $\begingroup$ 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 $V^{\ast}$. $\endgroup$
    – nfdc23
    Jan 3, 2016 at 4:14
  • $\begingroup$ 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 $\mathbf{R} \otimes_{\mathbf{Z}} L = V$. $\endgroup$
    – nfdc23
    Jan 3, 2016 at 4:15
  • 2
    $\begingroup$ 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). $\endgroup$
    – nfdc23
    Jan 3, 2016 at 4:24
  • $\begingroup$ @nfdc23 nice argument to show discreteness, thanks! I don't understand your hint about how to show $\mathbb{R}$-span of $L^*$ is $V^*$. $\endgroup$
    – usr0192
    Jan 3, 2016 at 6:55
  • $\begingroup$ I have written an extended answer below to explain what my hint was trying to get at. $\endgroup$
    – nfdc23
    Jan 4, 2016 at 4:24

1 Answer 1


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 $V^{\ast}$!).

The answer below is rather longer than the actual content, since we wish to explain why the relevant formulas don't just come out of thin air. If you don't care about motivation, just go immediately to the final paragraph for a quick self-contained proof.

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.


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