# Solving a (fun!) coequalizer problem for $\mathrm{SL}_n(\mathbb{R})\rightarrow\mathrm{SL}_n(\mathbb{C})$ in $\mathbf{Grp}$

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.

• 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. – Kevin Arlin Oct 21 at 5:17
• The opposite of a monoid (regarded as a category) is still a monoid, right? So $f,g$ are contravariant functors. – Fosco Oct 21 at 7:31
• What is $|A|$ ? Is it the determinant ? If so, $|\frac{A}{|A|}| = \frac{|A|}{|A|^n}$ which is hardly ever $1$ – Maxime Ramzi Oct 21 at 7:44
• @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 $\mathrm{SL}_n(\mathbb{R})$. – Kevin P. Barry Oct 21 at 9:05
• @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 $g$. – Kevin P. Barry Oct 21 at 9:20

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.

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.

• 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. – Servaes Oct 21 at 19:37
• 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? – Kevin P. Barry Oct 22 at 0:38
• @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. – Qiaochu Yuan Oct 22 at 3:48
• @Kevin: yes, it follows pretty easily, and yes, if you know all the normal subgroups of $G$ then this tells you all possible coequalizers. – Qiaochu Yuan Oct 22 at 3:49
• @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. – Servaes Oct 22 at 8:00