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.

  • $\begingroup$ Your claim that $\forall x,y \in A.\,x \otimes y = y \otimes x$ holds in $A \otimes A$ only if $A$ is cyclic is not correct. See mathoverflow.net/questions/119689 $\endgroup$ – HeinrichD Oct 11 '16 at 8:40
  • $\begingroup$ If A is an cyclic abelian group then $\tau=id$, right? But the other way around is wrong, you are right. I am trying to read your link. I think that the argument to apply is that in the first four lines of the first answer (by Will Sawin). I am trying to figure it out. $\endgroup$ – marcos Oct 11 '16 at 9:55
  • $\begingroup$ The simplest counterexample is $A=\mathbb{Z}[1/2]$. Here, $2$ is invertible and $\forall x,y \in A.\,x \otimes y = y \otimes x$ holds. Thus, $\langle x \otimes y + y \otimes x : x,y \in A \rangle = A \otimes A$. $\endgroup$ – HeinrichD Oct 11 '16 at 10:15
  • $\begingroup$ By $\mathbb{Z}[1/2]$ you mean the extension by 1/2 of the ring of integers right? $\endgroup$ – marcos Oct 11 '16 at 15:47
  • $\begingroup$ $B$ is the image of the linear endomorphism of $A\otimes A$ given as $(x\otimes y)\mapsto x\otimes y- \alpha(y)\otimes x$. You're asking about its surjectivity, and $B$ is finite, so this is also equivalent to its injectivity. $\endgroup$ – YCor Oct 11 '16 at 16:21

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

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 $\det(b_T)\neq 0$.

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 $p$.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.