In $PGL(n,\mathbb{R})$, is every real element strongly real?

Let $A\in GL(3,\mathbb R)$ and let $[A]\in PGL(3,\mathbb R)$ be its projective class. Assume $[A]$ is real in $PGL(3,\mathbb R)$, i.e. $[A]$ is conjugate to its inverse $[A^{-1}]$. Geometrically (via conics), one can prove that there exists an involution $[P]\in PGL(3,\mathbb R)$ with $$ [P][A][P]^{-1}=[A^{-1}]. $$ I am asking for a direct proof of this, avoiding the conic $\leftrightarrow \mathbb P^1$ detour in the proof below.

Definitions

• An element $g$ of a group $G$ is called real if it is conjugate to its inverse: $\exists h\in G$ with $hgh^{-1}=g^{-1}$.
• An element $g$ is called strongly real if it is conjugate to its inverse by an involution: $\exists h\in G$ with $h^2=1$ and $hgh^{-1}=g^{-1}$.

Basic facts about “strongly real”

1. Every involution is strongly real.
   If $g^2=1$ then $g=g^{-1}$, so $g$ is conjugate to $g^{-1}$ by the identity involution.

2. Every product of two involutions is strongly real.
   If $g=ab$ with $a^2=b^2=1$, then $$ aga^{-1}=a(ab)a=ba=g^{-1}. $$ So $a$ (an involution) conjugates $g$ to $g^{-1}$.

3. Conversely, every strongly real element is a product of two involutions.
   If $aga^{-1}=g^{-1}$ for some involution $a$, then $$ g=a(ag), $$ and $(ag)^2=1$. Hence $g$ is a product of involutions.

Thus: $$ g \text{ is strongly real}\quad\Longleftrightarrow\quad g \text{ is a product of two involutions}. $$

The geometric proof that a real class in $PGL(3,\mathbb R)$ is strongly real

Let $V=\mathbb R^3$. We identify projective transformations with classes $[A]\in PGL(V)$, where $A\in GL(V)$.

Step 1. Conics fixed by $[A]$ are eigenvectors of the pullback action on quadratic forms

A (real) conic in $\mathbb P(V)$ is the zero-locus of a nonzero quadratic form $q(x)=x^\top M x$ with $M=M^\top$ defined up to scale.
Let $S^2 V^\ast$ denote the 6-dimensional space of quadratic forms.
$A$ acts on $S^2 V^\ast$ by pullback: $$ (A\cdot q)(x)=q(Ax)\quad\Longleftrightarrow\quad A^\top M A = M' \text{ (matrix level)}. $$ Thus a conic defined by $M$ is projectively fixed by $[A]$ iff $$ A^\top M A = \lambda M \qquad(\lambda\neq 0), $$ i.e. $M$ is an eigenvector of the linear operator $T_{A}:S^2 V^\ast\to S^2 V^\ast$, $M\mapsto A^\top M A$.

Step 2. Existence of a nondegenerate quadratic form in the eigenspace of $T_A$

This step relies on the equivalence in this question.

We keep $V=\mathbb R^3$, $T_A:M\mapsto A^\top M A$ on $S^2V^\ast$, and write the eigenvalues of $A$ (over $\mathbb C$) as $\lambda_1,\lambda_2,\lambda_3$ with algebraic multiplicity.
From $[A]\sim[A^{-1}]$ in $PGL(3)$ we know there is a scalar $\mu\ne 0$ such that the multisets $\{\lambda_i\}$ and $\{\mu\,\lambda_i^{-1}\}$ coincide. Consequently $\mu$ is an eigenvalue of $T_A$, and—crucially—coincidences occur among the weights $\lambda_i\lambda_j$.

A convenient parametrization is: there exist $a\in\mathbb C^\times$ and $r\in\mathbb C^\times$ with $$ \{\lambda_1,\lambda_2,\lambda_3\}=\{a,\ ar,\ ar^2\} \qquad\text{and}\qquad \mu=a^2 r, $$ so that the relation $\lambda_2^2=\lambda_1\lambda_3$ holds (after reindexing). Note that either $r\neq 1$ (all eigenvalues distinct) or $r=1$ (all equal). There is no “two equal, one different” possibility compatible with $\lambda_2^2=\lambda_1\lambda_3$.

We now treat these two regimes separately. The goal is to exhibit, inside the $\mu$–eigenspace of $T_A$, either

• a nondegenerate real symmetric eigenvector $M$ (yielding a smooth fixed conic), or
• when that subspace consists only of degenerate forms, a direct linear-algebra construction of an involution $P$ with $PAP^{-1}=\text{(scalar)}\cdot A^{-1}$ (hence $[P][A][P]^{-1}=[A^{-1}]$).

Case I: $r\neq 1$ (three distinct eigenvalues) — diagonalizable case

Over $\mathbb C$, choose an eigenbasis $x_1,x_2,x_3$ for $A$. Then $T_A$ acts diagonally on the monomials: $$ T_A(x_i^2)=(\lambda_i^2)\,x_i^2,\qquad T_A(x_ix_j)=(\lambda_i\lambda_j)\,x_ix_j. $$ With $\mu=a^2r$ we have $$ T_A(x_2^2)=\lambda_2^2 x_2^2 \quad\text{and}\quad T_A(x_1x_3)=\lambda_1\lambda_3\, x_1x_3, $$ and $\lambda_2^2=\lambda_1\lambda_3=\mu$, so both $x_2^2$ and $x_1x_3$ lie in the $\mu$–eigenspace. Thus $$ E_\mu \supset \mathrm{Span}\{x_2^2,\ x_1x_3\}, $$ a 2-dimensional subspace. Any real linear combination $$ Q=\alpha\,x_2^2+\beta\,x_1x_3\qquad(\beta\ne 0) $$ is indefinite (take $(x_1,x_3)=(1,\pm1)$, $x_2=0$ to see $Q$ takes both signs), hence represents a nondegenerate real symmetric matrix $$ M=\begin{bmatrix} 0 & 0 & \tfrac{\beta}{2}\\ 0 & \alpha & 0\\ \tfrac{\beta}{2} & 0 & 0 \end{bmatrix}\!, \quad\det M=-\tfrac{1}{4}\alpha\beta^2\ne 0\ \text{ for generic }(\alpha,\beta). $$ So $A^\top M A=\mu M$ with $M$ nondegenerate: $[A]$ fixes the smooth real conic $\{x^\top Mx=0\}$.

This recovers the “eigenvalue-coincidence $\Rightarrow$ 2D eigenspace $\Rightarrow$ nondegenerate conic” mechanism in the diagonalizable setting.

Case II: $r=1$ (all eigenvalues equal) — unipotent possibilities

Write $A=\lambda U$ with $U$ unipotent. In $PGL$, the scalar $\lambda$ is irrelevant, and $$ A^\top M A=\mu M\quad\Longleftrightarrow\quad U^\top M U = M $$ since $\mu=\lambda^2$. There are two Jordan types for $U$ in dimension $3$:

• (II.a) A single Jordan block $J_3(1)$.
  Let $U=I+N$ with $$ N=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix},\quad N^3=0. $$ Solving $(I+N)^\top M (I+N)=M$ with $M=M^\top$ gives the solution space $$ M=\begin{bmatrix} 0 & 0 & 2a\\ 0 & -2a & a\\ 2a & a & b \end{bmatrix}\qquad(a,b\in\mathbb R), $$ which is 2-dimensional. Its determinant is $\det M=8a^3$, so for $a\ne 0$ these are nondegenerate and, as before, indefinite (hence define smooth real conics). Thus even in this non-diagonalizable case the $\mu$–eigenspace of $T_A$ contains nondegenerate conics.

• (II.b) A size $2+1$ Jordan form $J_2(1)\oplus J_1(1)$.
  In a Jordan basis take $$ U=I+N,\qquad N=\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix},\quad N^2=0. $$ Solving $(I+N)^\top M (I+N)=M$ now yields $$ M=\begin{bmatrix} 0 & 0 & 0\\ 0 & * & *\\ 0 & * & * \end{bmatrix}, $$ i.e. the entire $1$-eigenspace of $T_U$ consists of degenerate forms (the first row/column vanish), so this route does not produce a smooth fixed conic.

  However, in this case we can directly produce the required involution conjugator purely linearly: set $$ P=\mathrm{diag}(-1,1,1)\ \ \text{(an involution).} $$ Then $$ PNP^{-1}=-N,\qquad\text{so}\qquad PUP^{-1}=P(I+N)P^{-1}=I-N=U^{-1}, $$ and hence $$ PAP^{-1}=\lambda\,U^{-1}=\lambda\,(I-N) = \lambda^2\,(I+N)^{-1} = \lambda^2\,A^{-1}. $$ Projectively, this says $[P][A][P]^{-1}=[A^{-1}]$ with $[P]$ an involution. Thus even without a smooth fixed conic, the element is strongly real.

Takeaway

• If $[A]\sim[A^{-1}]$ and the eigenvalues are in geometric progression with ratio $r\ne 1$, the coincident weights for $T_A$ produce a 2D $\mu$–eigenspace containing indefinite, nondegenerate forms (hence a smooth fixed conic).
• If all eigenvalues are equal ($r=1$), then
  • for the $J_3(1)$ unipotent, the same conclusion holds by an explicit 2D family of nondegenerate solutions to $U^\top M U=M$;
  • for the $J_2(1)\oplus J_1(1)$ unipotent, the $\mu$–eigenspace consists of degenerate forms only, but we can still construct an involution $P$ explicitly with $[P][A][P]^{-1}=[A^{-1}]$.

Step 3. The stabilizer of a smooth real conic is $PGL(2,\mathbb R)$

Fix such a conic $C$. The quadratic Veronese embedding $$ \nu_2:\mathbb P^1 \longrightarrow \mathbb P^2,\qquad [u:v]\longmapsto [u^2:uv:v^2], $$ identifies $\mathbb P^1(\mathbb R)$ with $C(\mathbb R)$ after a projective change of coordinates, and it identifies the conic stabilizer with $PGL(2,\mathbb R)$: $$ \operatorname{Stab}(C)\ \cong\ PGL(2,\mathbb R). $$ Concretely, the homomorphism $\rho:PGL(2)\to PGL(3)$ is induced by the degree-2 action on binary quadratics; in matrices (choosing $C$ to be $x_1^2+x_2^2-x_3^2=0$) one has $$ \rho(a)^\top\, J\, \rho(a) \;=\; (\det a)^2\, J\qquad\big(J=\mathrm{diag}(1,1,-1)\big), $$ so $\rho(a)$ preserves $C$ projectively.

Step 4. Inside $PGL(2)$, inversion is realized by an involution

Let $$ J_2=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\in GL(2). $$ A direct $2\times2$ computation gives, for every $a\in GL(2)$, $$ J_2\, a\, J_2^{-1} \;=\; (\det a)\, a^{-T}. $$ Passing to projective classes, $$ [J_2]\,[a]\,[J_2]^{-1} \;=\; [a^{-1}]\qquad\text{in }PGL(2). $$ Moreover, $J_2^2=-I_2$, so $[J_2]$ has order $2$ in $PGL(2)$, i.e. $[J_2]$ is an involution.

Transporting via $\rho$, define $$ P_C \ :=\ \rho(J_2)\in PGL(3). $$ Then for every $[B]\in\operatorname{Stab}(C)$, $$ [P_C]\,[B]\,[P_C]^{-1} \;=\; [B^{-1}]\qquad\text{in }PGL(3), $$ and $[P_C]$ is an involution (since $J_2^2=-I_2$ maps to the identity class).

Step 5. Conclude for $[A]$

From Step 2, after conjugating $[A]$ inside $PGL(3,\mathbb R)$ we may assume $[A]\in\operatorname{Stab}(C)$.
Apply Step 4 inside $\operatorname{Stab}(C)$: there is an involution $[P_C]$ with $[P_C][A][P_C]^{-1}=[A^{-1}]$.
Undo the conjugation: if $[Q][A][Q]^{-1}\in\operatorname{Stab}(C)$, then $$ [P]\ :=\ [Q]^{-1}\,[P_C]\,[Q]\quad\text{is an involution and}\quad [P][A][P]^{-1}=[A^{-1}]. $$

Hence every real class $[A]\in PGL(3,\mathbb R)$ is strongly real, and the conjugating element can be taken to be an involution.

Question

The geometric proof above uses that

• $[A]\sim[A^{-1}] \iff$ $[A]$ preserves a smooth conic $C$ (Step 2),
• the conic stabilizer is $PGL(2)$ (Step 3),
• in $PGL(2)$ the class of $J_2$ is an involution that conjugates any element to its inverse (Step 4).

All these steps can be phrased with coordinate matrices, but they conceptually route through the identification of $C$ with $\mathbb P^1$.

Request. Can one prove directly, that if $[A]\sim[A^{-1}]$ in $PGL(3,\mathbb R)$ then there exists an involution $[P]$ with $[P][A][P]^{-1}=[A^{-1}]$, without invoking conics or the $PGL(2)$–stabilizer identification?

For context, in the same spirit:

• $n=1$. Here $GL(1,\mathbb R)=\mathbb R^\times$, and we mod out by the scalar group $\mathbb R^\times$ again, so $$ PGL(1,\mathbb R)=\mathbb R^\times/\mathbb R^\times\cong\{1\}. $$ In particular, $PGL(1,\mathbb R)$ is the trivial group. The unique element is vacuously real and strongly real.

• $n=2$.
  Work in $GL(2,\mathbb R)$ with $$ J_2=\begin{bmatrix}0&1\\-1&0\end{bmatrix},\qquad J_2^{-1}=-J_2. $$ For $a=\begin{bmatrix}p&q\\ r&s\end{bmatrix}$, compute $$ J_2\,a=\begin{bmatrix}r&s\\ -p&-q\end{bmatrix},\qquad (J_2\,a)\,J_2^{-1}=\begin{bmatrix}s&-r\\ -q&p\end{bmatrix}=\operatorname{adj}(a). $$ Since $\operatorname{adj}(a)=(\det a)\,a^{-T}$, we obtain the identity $$ J_2\,a\,J_2^{-1}=(\det a)\,a^{-T}. $$ Passing to projective classes kills the scalar factor $\det a$, hence in $PGL(2,\mathbb R)$ $$ [J_2]\,[a]\,[J_2]^{-1}=[a^{-1}]. $$ Finally, $J_2^2=-I_2$, so $[J_2]$ has order $2$ in $PGL(2)$, i.e. it is an involution.
  Conclusion: every element of $PGL(2,\mathbb R)$ is conjugate to its inverse by the involution $[J_2]$, so every element is strongly real in $PGL(2,\mathbb R)$.

• $n=3$. As shown above, every real element is strongly real.

What about $n\geq 4$? Is the implication “real $\implies$ strongly real” always valid in $PGL(n,\mathbb R)$, or do counterexamples exist in higher dimensions?

General question. For arbitrary $n$, in $PGL(n,\mathbb R)$: if $[A]$ is real (i.e. $[A]\sim[A^{-1}]$), must it also be strongly real (i.e. conjugate to $[A^{-1}]$ by an involution)?