Subgroups of the tensor product $A\otimes A$

Question

I have this problem about subgroups of the tensor product of an abelian group $A$ with itself which arises from a complete different setting. I fell into this question studying quandles and quandle coverings, but the question can be stated just in terms of abelian groups.

Let $A$ be an Abelian group and $\alpha$ an automorphism of $A$ such that $1-\alpha$ is invertible.

I need to compute the following subgroups in $A\otimes A$:

$B=\langle x\otimes y-\alpha(y)\otimes x,\quad x,y\in A\rangle$

The goal is to find some condition on $A$ and $\alpha$ in order to have that $B=A\otimes A$. For instance, if $A$ is a cyclic group then $B=A\otimes A$. Moreover, I know that $B$ is the whole group when $A$ is elementary abelian different from $\mathbb{Z}_2\times \mathbb{Z}_2$ and $\alpha$ has order $|A|-1$ (it follows by an equivalent property in quandles setting).

A counterexample is given when $\alpha=-1$. Since $B=\langle x\otimes y+y\otimes x,\quad x,y\in A\rangle$ is a subgroup of the element fixed by the flip $\tau$ ($\tau(x\otimes y)=y\otimes x$) and then if $A$ is not cyclic, this is a proper subgroup.

Thank you. M.

P.S. I forgot to say that $A$ is finite.

Answer 1

Here's a full answer. First I consider a general linear algebra problem, then I apply it to the specific given problem.

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Let $K$ be a field, $V$ a finite-dimensional vector space, $T$ an operator of $V$. Consider the endomorphism $b_T$ of $V\otimes V$ (all $\otimes$ are over $K$) given by $$x\otimes y\mapsto b_T(x\otimes y)=x\otimes y-Ty\otimes x.$$ Then I claim that $b_T$ is an automorphism (or equivalently surjective) if and only if all the following hold:

(a) 1 is not eigenvalue of $T$;

(b) $-1$ is not double eigenvalue of $T$ [that is, $\dim(\mathrm{Ker}(1+T))\le 1$];

(c) $T$ has no pair of distinct inverse eigenvalues (in any extension field of $K$).

Since both the problem and the characterization ignore field extension, it is enough to prove this when $K$ is algebraically closed; we do it by induction on $d=\dim(V)$; the case $d=0$ being trivial, assume $d\ge 1$. Write $b_T=b_{V,T}$. Fix an eigenline $E\subset V$ of $T$. It yields a $b_T$-stable $(2d-1)$-dimensional subspace $W=(E\otimes V+V\otimes E)$ of $V\otimes V$. So $\det(b_T)=\det((b_T)|_{W})\det(b_{V/E,T})$, because the endomorphism of $(V\otimes V)/W$ induced by $T$ can be identified with $b_{V/E,T}$. So we have to compute this first determinant $\det((b_T)|_{W})$. Let $(e_1,\dots,e_d)$ be a basis of $V$ with $e_1\in E$ and $Te_1=\lambda e_1$. Then a basis of $W$ is given by $$(e_1\otimes e_1,e_2\otimes e_1,\dots e_d\otimes e_1,e_1\otimes e_2,\dots e_1\otimes e_d).$$ Writing matrices by blocks $1+(d-1)+(d-1)$, we see that the matrix of $(b_T)|_{W}$ in this basis is $$\begin{pmatrix}1-\lambda & 0 & *\\ 0 & I & -T'\\0 & -\lambda I & I\end{pmatrix},$$
where $T'$ is the matrix of the endomorphism induced by $T$ on $V/E$ on the basis $(e_2,\dots,e_d)$. The determinant of $(b_T)|_{W}$ is therefore $(1-\lambda)\det(1-\lambda T')$.

Thus the determinant of $b_T$ is $$\det(b_T)=(1-\lambda)\det(1-\lambda T')\det(b_{V/E,T}).$$

From this formula, we can deduce the result. First, for the easy implication: if $1$ is eigenvalue as in (a), pick $\lambda=1$ and then $\det(b_T)=0$. If there are two inverse eigenvalues with independent eigenvectors (this covers (b) and (c)), then pick $\lambda$ to be one of them, and then since $\lambda^{-1}$ is an eigenvalue of $T'$, we have $\det(1-\lambda T')=0$ and hence $\det(b_T)=0$.

Conversely, assume that none of (i),(ii),(iii) holds. By induction and using (i)-(ii)-(iii), $\det(b_{V/E,T})\neq 0$. Also $\det(1-\lambda T')\neq 0$: this is clear if $\lambda=0$, and if $\lambda\neq 0$ we know from (ii),(iii) that $\lambda^{-1}$ is not eigenvalue of $T'$. Finally $\lambda\neq 1$ by (i) and hence $\det(b_T)\neq 0$.

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Now consider the analogous problem in a finite abelian group $A$. An endomorphism of a finite abelian group $Q$ is surjective if and only if the induced endomorphism of $Q/pQ$ is surjective for every prime $p$. Noting that $(A/pA\otimes A/pA)=(A\otimes A)/p(A\otimes A))$ and that this "commutes" with defining the operator $b_T$, we thus reduce all the problem to the case when $A$ is a $p$-elementary abelian group, in which case it is solved by the above linear algebra problem (with $K=\mathbf{Z}/p\mathbf{Z}$).

To summarize, given an endomorphism $T$ of $A$, the endomorphism of $A\otimes A$ given by $b_T(x\otimes y)=x\otimes y-Ty\otimes x$ is surjective if and only if the three following condition hold, denoting $T_p$ the endomorphism of $T$ induced on $A/pA$:

(a') for every prime $p$, 1 is not eigenvalue of $T_p$;

(b') for every prime $p$, $-1$ is not double eigenvalue of $T_p$ [that is, $\dim_{\mathbf{Z}/p\mathbf{Z}}(\mathrm{Ker}(1-T_p))\le 1$];

(c') for every prime $p$, $T_p$ has no pair of distinct inverse eigenvalues (in any extension field of $\mathbf{Z}/p\mathbf{Z}$).

These conditions can sometimes be restated: (a') just means that $1-T$ is invertible (which was one of your assumptions). Actually, in the linear setting, given (a), (b)-(c) means that 1 is not an eigenvalue of the endomorphism $T\wedge T$ of the exterior product $V\wedge V$. So probably given (a'), (b')-(c') means that the $1-T\wedge T$ is an invertible endomorphism of $\Lambda^2 A=(A\otimes A)/\langle a\otimes a:a\in A\rangle$. Note that in spite of the latter characterization being shorter, it is much more convenient in practice to reduce modulo $p$.