On tensor products of semistable lattices

If $V$ and $W$ are Euclidean spaces (finite dimensional inner product spaces over $\mathbb{R})$ then so is $V\otimes W$ with the inner product

If $\left\{e_{i}\right\}$ is an orthonormal basis of $V$ and $\left\{e_{j}^{\prime}\right\}$ an orthonormal basis of $W$ then $\left\{e_{i}\otimes e_{j}^{\prime}\right\}$ is an orthonormal basis of $V\otimes W.$ In this way $V^{\otimes k}$ is endowed with a Euclidean structure.

The $k$-th exterior power $\bigwedge^{k}V$ is the quotient of $V^{\otimes k}$ by the subspace $\mathcal{N}$ spanned by tensors $v_{1}\otimes\cdot\cdot\cdot\otimes v_{k}$ in which, for some $i\neq j$, $v_{i}=v_{j}.$ The orthogonal complement $\mathcal{N}^{\perp}$ is the space of alternating $k$-tensors, spanned by

and this tensor projects modulo $\mathcal{N}$ to $v_{1}\wedge\cdot\cdot\cdot\wedge v_{k},$ the image modulo $\mathcal{N}$ of $v_{1}\otimes\cdot\cdot\cdot\otimes v_{k}.$ We may therefore identify $\bigwedge^{k}V$ with $\mathcal{N}^{\perp}$. As such it inherits from $V^{\otimes k}$ a Euclidean structure. If $\left\{e_{i}\right\}$ is an orthonormal basis of $V$ then $e_{I}=e_{i_{1}}\wedge\cdot\cdot\cdot\wedge e_{i_{k}},$ for $I=(i_{1}<\dots<i_{k}),$ is a basis of $\bigwedge^{k}V,$ which is orthogonal but not normalized: the norm of $e_{I}$ is $1/\sqrt{k!}.$ To correct it, we modify the inner product that $\bigwedge^{k}V$ inherits from $V^{\otimes k}$ by a factor of $k!$ and set

In this inner product the $e_{I}$ form an orthonormal basis. Moreover, the volume of the $k$-dimensional parallelopiped spanned by $v_{1},\dots,v_{k}$ in $V$ is nothing but the norm $|v_{1}\wedge\cdot\cdot\cdot\wedge v_{k}|.$ To see it, choose an orthonormal basis $e_{i}$ of $V$ the first $k$ vectors of which span $Span\left\{v_{1},\dots,v_{k}\right\}$ (assuming the $v_{i}$ are linearly independent) and observe that

where $v_{i}=\sum a_{ij}e_{j}.$

To begin the proof, note that

If $\alpha$ is a vector in $L\otimes M$ we denote by $l(\alpha)$ the length of $\alpha,$ which is the minimal number $l$ such that

(To avoid abuse of language, $|\alpha|$ will always be called the norm of $\alpha$, and not its length.) In such a case, the $u_{k}$ and the $u_{k}^{\prime}$ are linearly independent vectors in $L$ and $M$ respectively. Thus the length of a vector is at most $\min(rk(L),rk(M)).$

Let us write, for any lattice $M$

The maximum is over all sublattices of all ranks (it is easily seen that the maximum is attained). A unimodular lattice $M$ is semistable if and only if $\mu_{\max}(M)=0.$

The proposition proves the case $\mathbf{rk(S)=1}$ of the theorem, but to prove the cases $rk(S)=2$ or $3$ we shall need its full strength.

Let now $S$ be a sublattice of rank $2$ in $L\otimes M.$ Let $\alpha$ be a nonzero vector of shortest norm in $S,$ and $\beta$ another vector in $S$ such that $S=\mathbb{Z}\alpha+\mathbb{Z}\beta$. Subtracting a multiple of $\alpha$ from $\beta$ we may assume that the angle between $\alpha$ and $\beta$ is between $60^{\circ}$ and $120^{\circ}.$ It follows that

If either of $\alpha$ or $\beta$ is indecomposable (of length $\geq 2$) we are done, since its norm is then at least $\sqrt{2}$ and the other’s norm is at least 1, but $\sqrt{2}\sqrt{3}>2.$ We may therefore assume that

are both decomposable tensors. We now use the identity

which follows at once from the fact that

If $v_{1}$ and $v_{2}$ are proportional, then $w_{1}$ and $w_{2}$ are independent, and $|v_{1}|^{2}|v_{2}|^{2}|w_{1}\wedge w_{2}|^{2}$ $\geq 1,$ while the second and the third terms vanish. Otherwise, the first term is larger than the third, and the second is $\geq 1$ by our assumption. This concludes the rank 2 case.

The rank 3 case is handled similarly, but the details are more complicated. Let $rk(S)=3.$ Let $\alpha$ be a shortest nonzero vector in $S$ and $\beta$ a second shortest vector independent from $\alpha,$ and assume as before that the angle between $\alpha$ and $\beta$ is between $60^{\circ}$ and $120^{\circ}.$ Complete to a basis of $S$ by a vector $\gamma,$ $|\gamma|\geq|\beta|\geq|\alpha|\geq 1.$ Using orthonormal coordinates $x,y,z$ in the real subspace spanned by $S,$ such that the $x$-axis is in the $\alpha$-direction and the $(x,y)$-plane is the plane spanned by $\alpha$ and $\beta,$ we may assume, subtracting from $\gamma$ a suitable integral linear combination of $\alpha$ and $\beta$ that

with $|x|\leq|\alpha|/2$ and $|y|\leq|\beta|/2.$ This is because the rectangle

where $\theta$ is the angle between $\alpha$ and $\beta,$ is a fundamental domain for $\mathbb{Z}\alpha+\mathbb{Z}\beta.$ We now have

and

(by the rank 2 case), so

If $l(\alpha)$, $l(\beta)$ or $l(\gamma)\geq 2,$ then by the proposition and the fact that $|\gamma|\geq|\beta|\geq|\alpha|,$ we must have $|\gamma|^{2}\geq 2,$ and we are done.

There remains the case $\alpha=v_{1}\otimes w_{1},$ $\beta=v_{2}\otimes w_{2}$ and $\gamma=v_{3}\otimes w_{3},$ in which the proposition does not help us. The key in this case is the following identity.

We continue the proof of the theorem, when $\alpha$, $\beta$ and $\gamma$ are decomposable as above. Choose an orthonormal basis $e_{1},e_{2},e_{3},...$ of $L_{\mathbb{R}}$ so that $v_{1}\in\langle e_{1}\rangle,$ $v_{2}\in\langle e_{1},e_{2}\rangle,$ $v_{3}\in\langle e_{1},e_{2},e_{3}\rangle$ and similarly an orthonormal basis $e_{1}^{\prime},e_{2}^{\prime},e_{3}^{\prime},...$ of $M_{\mathbb{R}}$ putting the $w_{i}$ in a triangular form. Assume first that the $v_{i}$ are linearly independent. Expanding in the orthonormal basis $e_{i}\otimes e_{j}^{\prime}$ of $L_{\mathbb{R}}\otimes M_{\mathbb{R}}$ and in the corresponding orthonormal basis of of $\bigwedge^{3}(L_{\mathbb{R}}\otimes M_{\mathbb{R}})$ (recall the normalization from 1.1)

Assume next that both the $v_{i}$ and the $w_{i}$ are linearly dependent. If $v_{1}$ and $v_{2}$ are proportional then $w_{1}$ and $w_{2}$ can not be proportional, otherwise $rk(S)\leq 2.$ In this case we may assume that $v_{1},v_{2}\in\langle e_{1}\rangle$ and $v_{3}\in\langle e_{1},e_{2}\rangle,$ while $w_{1}\in\langle e_{1}^{\prime}\rangle$ and $w_{2},w_{3}\in\langle e_{1}^{\prime},e_{2}^{\prime}\rangle.$ We find

There remains the case where the $v_{i}$ and the $w_{i}$ are linearly dependent, but no two vectors in each triplet are proportional.

Introduce the notation

and similarly $\mu_{i}=|w_{i}||w_{i^{\prime}}\wedge w_{i^{\prime\prime}}|.$ Note that

where $\theta_{i}$ is the angle between $v_{i^{\prime}}$ and $v_{i^{\prime\prime}}$ in the plane spanned by the $v_{i}^{\prime}s.$ We may assume that $\theta_{i}=\theta_{i^{\prime}}+\theta_{i^{\prime\prime}}.$ A direct computation shows then that

Similarly denote by $\omega_{i}$ the angle between $w_{i^{\prime}}$ and $w_{i^{\prime\prime}}.$ We now observe that in the 18-term relation, nine terms drop out and the remaining nine give

This concludes the proof of the theorem.

Duality allows us to conclude more. Quite generally, if

is an exact sequence of Euclidean lattices (this means that $Q/P$ is torsion-free and that the metrics on $P_{\mathbb{R}}$ and $R_{\mathbb{R}}$ are the ones induced from the metric on $Q_{\mathbb{R}}$) then

If $R^{\vee}$ is the dual lattice, $\mu(R^{\vee})=-\mu(R).$ From these two observations it follows at once that if $L$ is semistable, so is $L^{\vee}.$

Let $L$ and $M$ be two unimodular semistable lattices of ranks $n$ and $m$. Then $L^{\vee}$ and $M^{\vee}$ are unimodular and semistable too. Suppose we know the desired results for all $S$ of some rank $s.$ Let $P$ be a rank $nm-s$ (we shall say it has corank $s$) sublattice of $L\otimes M$ for which we want to prove that $\mu(P)\leq 0.$ We may assume that $\,(L\otimes M)/P$ is torsion free. We equip it with the quotient metric and call it $R$. Then $R$ is a rank $s$ lattice and $R^{\vee}=P^{\perp}$ is a rank $s$ sublattice of $L^{\vee}\otimes M^{\vee}.$ By what we have seen already, $\mu(R^{\vee})\leq 0,$ but

since $\mu(L\otimes M)=0,$ and $\mu(R)=-\mu(R^{\vee}).$ This shows that if we know the desired result for all rank $s$ lattices, we also know it for all corank $s$ lattices.

[Fa] Faltings, G.: Mumford-Stabilität in der algebraischen Geometrie, Proceedings of the 1994 ICM.

[To] Totaro, B.: Tensor products in $p$-adic Hodge theory, Duke Math.J. 83 (1996), 79-104.