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First off, the problem posed below is mostly arbitrary; it's just for my own education. (And maybe for yours, as well.)

It's fairly clear to me what the (co)equalizers of abelian groups in $\mathbf{Grp}$ are, but it's less clear what those mean for non-abelian groups. So, I came up with a problem that seems non-trivial and interesting.


I'm trying to coequalize $f,g:\mathrm{SL}_n(\mathbb{R})\rightrightarrows\mathrm{SL}_n(\mathbb{C})$, where

  • $f(A)=A$
  • $g(A)=(A^*)^{-1}$

(Both purposely not surjective.)


To solve this, we need to find "the best" $l:\mathrm{SL}_n(\mathbb{C})\rightarrow L$. For now, I'll settle for any $L$ that isn't $\{0\}$.

The images of both $f$ and $g$ are $\mathrm{SL}_n(\mathbb{R})\subset\mathrm{SL}_n(\mathbb{C})$, so to start with I'll just look at that part of the domain of $l$.

  • $l(A^*)=l(A^{-1})$, based on $f$ and $g$. (Again, just on $\mathrm{SL}_n(\mathbb{R})$ for now.)
  • $l(AA^*)=l(A^*A)=e_L$, following from the statement above, and $l$ being a homomorphism.
  • Since $AA^*$ and $A^*A$ are positive-definite Hermitian (PDH), and PDH have Cholesky decompositions resembling $AA^*$, we can more generally say that $l(B)=e_L$ when $B$ is PDH. (Extending $l$ to $\mathrm{SL}_n(\mathbb{C})$.)
  • This also means that $l(D)=e_L$ when $D$ is diagonal with positive entries.
  • For any $A\in\mathrm{SL}_n(\mathbb{C})$, we can create an SVD $A=U\Sigma V^*$, with unitary $U$ and $V$, and $U,\Sigma,V\in\mathrm{SL}_n(\mathbb{C})$. Since $l(\Sigma)=e_L$, $l(A)=l(UV^*)$. ($UV^*$ should be unique, since $A$ is of full rank.)
  • If $A$ is unitary, it can be diagonalized as $A=VDV^*$ for unitary $V$ and diagonal $D$. Importantly, $D$ should only be in the kernel of $l$ if it only has positive (real) values, which is only true for $I$.

So it seems like $L$ is (at most) isomorphic to $\mathrm{SU}(n)$, with $l(A)$ taking $A$ to an equivalence class based on its rotation action after removing any distortion it makes. Does that sound accurate and/or reasonable? (For example, maybe a matrix with a non-real determinant can sneak in when removing $\Sigma$, thereby breaking $\mathrm{SL}_n(\mathbb{C})$.)


I spent several hours going through this, and I changed my conclusion about 5 times. The last few times were while proofreading. Whether or not my answer above is correct, I'd appreciate any pointers regarding shortcuts I could have taken, etc.

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    $\begingroup$ This isn’t that huge a deal, but $f$ and $g$ are anti-homomorphisms, right? They don’t technically lie in the category of groups. $\endgroup$ – Kevin Arlin Oct 21 at 5:17
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    $\begingroup$ The opposite of a monoid (regarded as a category) is still a monoid, right? So $f,g$ are contravariant functors. $\endgroup$ – Fosco Oct 21 at 7:31
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    $\begingroup$ What is $|A|$ ? Is it the determinant ? If so, $|\frac{A}{|A|}| = \frac{|A|}{|A|^n}$ which is hardly ever $1$ $\endgroup$ – Maxime Ramzi Oct 21 at 7:44
  • $\begingroup$ @MaximeRamzi Good point. In all my haste thinking about solving the problem, I totally forgot that that's not how you normalize a matrix. But, since I'm not doing an equalizer, I just changed the domain to $\mathrm{SL}_n(\mathbb{R})$. $\endgroup$ – Kevin P. Barry Oct 21 at 9:05
  • $\begingroup$ @KevinArlin I wasn't aware of antihomomorphisms! No wonder it was a mess when I tried to equalize $A$ and $A^*$ initially. I just moved the inverse from $f$ to $g$. $\endgroup$ – Kevin P. Barry Oct 21 at 9:20
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The coequalizer is trivial.

$SL_n(\mathbb{C})$ is almost a simple group (for $n \ge 2$, and it's trivial for $n = 1$): its center $Z(SL_n(\mathbb{C}))$ is the subgroup of scalar multiples of the identity where the scalar is an $n^{th}$ root of unity, and the quotient by the center is the projective special linear group $PSL_n(\mathbb{C})$, which is simple (either as an abstract group or as a Lie group; for simplicity as an abstract group see, for example, this note by Keith Conrad).

This implies that a normal subgroup containing any non-central element of $SL_n(\mathbb{C})$ must in fact be all of $SL_n(\mathbb{C})$, which is certainly the case for the normal subgroup describing this coequalizer.

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In general the coequalizer of $f,g:\ H \longrightarrow\ K$ in $\mathbf{Grp}$ is (isomorphic to) the quotient $K/N$ where $N$ is the normal closure of $$\{f(h)g(h)^{-1}:\ h\in H\}.\tag{1}$$ In this particular case we have $f(A)g^{-1}(A)=AA^{\ast}$, which is real symmetric and positive definite, so it orthogonally diagonalisable, i.e. $AA^{\ast}=Q_A^{-1}D_AQ_A$ for an orthogonal matrix $Q_A\in O_n(\Bbb{R})$ and a diagonal matrix $D_A\in\operatorname{SL}_n(\Bbb{R})$ with all diagonal entries positive. Note that we may also take $Q_A\in\operatorname{SL}_n(\Bbb{C})$ because a priori $\det Q_A=\pm1$, and hence also either either $$Q_A\in\operatorname{SL}_n(\Bbb{C})\qquad\text{ or }\qquad \zeta_{2n}Q_A\in\operatorname{SL}_n(\Bbb{C}),$$ where $\zeta_{2n}$ is a primitive $n$-th root of $-1$, and of course $\zeta_{2n}Q_A$ also satisfies $$(\zeta_{2n}Q_A)^{-1}D_A(\zeta_{2n}Q_A)=Q_A^{-1}D_AQ_A=AA^{\ast}.$$ Then the normalizer of $(1)$ contains \begin{eqnarray*} N&=&\{P^{-1}AA^{\ast}P:\ A\in\operatorname{SL}_n(\Bbb{R}),\ P\in\operatorname{SL}_n(\Bbb{C})\}.\\ &=&\{P^{-1}Q_A^{-1}D_AQ_AP:\ A\in\operatorname{SL}_n(\Bbb{R}),\ P\in\operatorname{SL}_n(\Bbb{C})\}.\\ &=&\{P^{-1}D_AP:\ A\in\operatorname{SL}_n(\Bbb{R}),\ P\in\operatorname{SL}_n(\Bbb{C})\}.\\ \end{eqnarray*} So the coequalizer is precisely the quotient of $\operatorname{SL}_n(\Bbb{C})$ by the normal subgroup generated by all diagonalizable matrices with real positive eigenvalues, which seems like a big subgroup. Apparently this is all of $\operatorname{SL}_n(\Bbb{C})$, and so the coequalizer is trivial.

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  • $\begingroup$ I posted this in a bit of a hurry because my battery will die in a minute. I will check the details in a few hours. $\endgroup$ – Servaes Oct 21 at 19:37
  • $\begingroup$ Thanks! Does this generalization easily follow from the axioms of $\mathbf{Grp}$? Also, does this imply that if one knows all of the normal subgroups of $G$ then one knows every potential coequalizer solution? $\endgroup$ – Kevin P. Barry Oct 22 at 0:38
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    $\begingroup$ @Servaes: the subset of diagonalizable matrices with real positive eigenvalues isn't a subgroup. $SL_n(\mathbb{C})$ is very close to being a simple group and it has very few normal subgroups. $\endgroup$ – Qiaochu Yuan Oct 22 at 3:48
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    $\begingroup$ @Kevin: yes, it follows pretty easily, and yes, if you know all the normal subgroups of $G$ then this tells you all possible coequalizers. $\endgroup$ – Qiaochu Yuan Oct 22 at 3:49
  • $\begingroup$ @QiaochuYuan You are absolutely right, I was a bit confused by the result myself when I posted this in a hurry. I have corrected this into a half-finished answer, as yours is more to the point anyway. $\endgroup$ – Servaes Oct 22 at 8:00

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