Classified at this point

$\langle|\rangle:M\times M\rightarrowtail \mathcal{O}$ where $$\langle u|v\rangle =\bar{u}^tDv^d$$

• Orthogonal: $\mathcal{O}$ field $K$, identity involution, $M=K^d$, $D$ diagonal, determinant square class defines the geometry.*
• Unitary: $\mathcal{O}=\left(\frac{\alpha}{K}\right)$ involution swaps roots (note may be split), $D=I_d$.
• Symplectic:$\mathcal{O}=\left(\frac{\alpha,\beta}{K}\right)$, adjugate involution, $M=(\mathcal{O}e)^{d}$ if split $e^2=e\not\in \{0,1\}$; else $M=\mathcal{O}^d$, $D=I_d$.
• Exceptional:$\mathcal{O}=\left(\frac{\alpha,\beta}{K}\right)$, $M=\mathcal{O}^{1,3}$, octonion/Albert involution, $D=I_d$.

*If $2K=0$ then orthogonal uses norm not product.

Start with Euclidean inner product

Unit circle Drawn

Fix some other

inner product

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...Gram-Schmidt should not be done in public...

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• Start with $e_1, e_2$
• $u_1=e_1$
• Replace $u_2=e_2-\frac{\langle e_2|u_1\rangle}{\langle u_1|u_1\rangle}u_1=\begin{bmatrix} -\frac{1}{2} \\ 1 \end{bmatrix}$

Now Find a Simultaneous

orthonormal basis for both!

Unit circles Drawn

...Gram-Schmidt should not be done in public...

Unit circle Drawn

• Start with $e_1, e_2$
• Normalize $e_1$ as $u=\frac{e_1}{\sqrt{\langle e_1|e_1\rangle}}=\begin{bmatrix} \frac{\sqrt{2}}{2}\\ 0 \end{bmatrix}$
• Replace $u_2=e_2-\frac{\langle e_2|u_1\rangle}{\langle u_1|u_1\rangle}u_1=\begin{bmatrix} -\frac{1}{2} \\ 1 \end{bmatrix}$
• Norm. $u_2=\frac{u_2}{\sqrt{\langle u_2|u_2\rangle}}=\begin{bmatrix} -\sqrt{2}/2\\\sqrt{2}\end{bmatrix}$

The Problem

•  ONE INNER PRODUCT: BE GREEDY.  Every nonzero vector $v$ is orthogonal to a hyperplane $H=v^{\bot}$, recurse.
• TWO INNER PRODUCTS:
  • $v$ is perpendicular to $H_1=v^{\bot(1)}$ and $H_2=v^{\bot(2)}$.
  • Only get an simultaneous ortho. basis if $H_1=H_2$.
• Generically $H_1\cap H_2$ has codimension 2, so not equal.
• Greed fails. You must know where to start.

Theorem. Ivanyos-Qiao

For $p\neq 2$ average case class 2 $p$-group isomorphism is in polytime.

Theorem. Ivanyos-Qiao

For $2K=K$ breaks the Patarin post-quantum cypher.

Theorem. Li-Qiao

For $2K=K$ solve isometry in polytime.

Theorems...Li-Qiao, Li, Qiao, et. al.

Complexity of tensor graph analogs.

Theorems. Grochow-Qiao

Group isomorphism by extension in polytime; tensor isomorphism, complixity classes, etc.

Brooksbank-Maglione-O'Brien-W.

Theorem. Lewis-W.

For $p\neq 2$ isomorphism of quotients of Heisenberg groups in polytime.

Find characteristic subgroups, isomorphism test, invariants.

Algorithm. Brooksbank-W.

For $p\neq 2$ adjoint-tensor isomorphism test.

Theorem. W.

The optimal bound on Gowers profiles is $\log n-2$.

Theorem. Brooksbank-Maglione-W.

Isomorphism testing genus 2 is polytime.

Theorem. Kassabov-Tybursky-W.

Subgroup lattices of simple groups are as diverse as possible.

Defn. W.

$\alpha$-Filter refinement.

$\alpha$-filters in polytime.

Theorem. Brooksbank-Li-Qiao-W.

A Weisefeller-Lehman for groups

Theorems.  Maglione

Outline

• $\mathrm{Isom}(\Phi) = \mathrm{Adj}(\Phi)^{\#}$
• Taft: $2K=K$ implies $(A,*) = (S,*)\ltimes (J(A),*)$
• Albert: $(S,*)=(S_1,*_1)\oplus \cdots\oplus (S_{\ell},*_{\ell})$ *-simples
• Weil: $(S_i,*_i)\cong \mathrm{Adj}(\phi_i:V_i\times V_i\rightarrowtail L_i)$
• Step I. Find ways to detect all this.
• Step II. write down $\bot_i$-bases of $V_i$.
• Step III. Write down Steinberg generators.
• Step IV. Glue along diagonal.
• Step V. Magic power series $z+\sqrt{1+z^2}$ a basis of $\mathrm{Alt}\cap J(A)$

Further Consequence

Precise structure formula $\forall \Phi,\exists \tau_1,\ldots,\tau_{\ell}$ where $$\begin{aligned} \mathrm{Isom}(\Phi)&=(\mathrm{Isom}(\tau_1)\times\cdots\times \mathrm{Isom}(\tau_{\ell}))\ltimes O_p\\ O_p & = \{z+\sqrt{1+z^2}\mid \exists n.(z^n=0)\wedge(z^*=-z)\}.\end{aligned}$$

Informally: Neo-classical geometry decomposes into a series of classical geometries.

$2K=K$

• Uni-triangular Isometries $$\{z+\sqrt{1+z^2}\mid \exists n.(z^n=0), z^*=-z\}$$
• Taft decomposition $$(A,*)=(S,*)\ltimes J(A)$$
• $(\mathbb{M}_d(K),*)=\mathrm{Sym}\oplus \mathrm{Alt}$
• $\mathrm{Isom}(\phi)=\mathrm{Adj}(\phi)^{\#}$

$2K=0$

• $z+\sqrt{1+z^2}$ divides by 0.
• $\exists (A,*)$ where $J(A)$ does not split.
• $\mathrm{Alt}\leq \mathrm{Sym}$
• Orthogonal groups are not unitary for ring involutions.

$z+\sqrt{1+z^2}$ asks too much

• Use a semi-standard induction through linear layers
• Actually you discover along the way that this induction has infinitely many cut-downs, no power series can exist.

Almost always Taft decompositions exist for subalgebras that contain $A^{\#}$

In non-orthogonal type this happens and the invariant is to commute with a derivation (1-cocycle).

How do show this? Assume so, write down what it would require, call all failures "exceptional".

The exceptional case

Over $L=K[t]/(t^2)$ in $\mathbb{M}_{2m}(L)$ make

No Taft splitting.

Isometries of $(A,*)/(J(A),*)$ do not lift.

Those that lift are

$$S^{\diamond}=\{s\in S\mid s^*s\equiv 1\mod{J(A)}, \exist m.((s+m)^*(s+m)=1)\}$$

These are quadratic equations, no hope to solve.

Reconfigure

After some algebra

$$\exists m.(\delta(s)= sm+ms)$$

$$S^{\diamond}=\{s\in S\mid sm\bar{s}=m\}$$

Still quadratic, but concentrates on a single defect, $m$.

Freshman's Dream

If $pK=0$ then $(a+b)^p=a^p+b^p$.

If $2K=0$ then for all $M\in \mathrm{Sym}$,

$$(A+B)M(A+B)^*\equiv AMA^*+BMB^*\pmod{\mathrm{Alt}}$$

Professor's Dream

Fix involution $M\mapsto M^*$ on $\mathbb{M}_d(K)$, 

$$\mathrm{Sym}=\{X\mid X^*=X\}\qquad \mathrm{Alt}=\{X\mid X^*=-X,\textnormal{0 diag}\}$$

Corollary. The isometries that lift are

$$\begin{aligned}S^{\diamond} & = \{s\mid \bar{s}s=1, sm\bar{s}=m\}\\ & = \{s\mid \bar{s}s=1, s\cdot m=m\}\\ & =\mathrm{Stab}_{\mathrm{Isom}(\phi)}(M)\end{aligned}$$

This is a known group: a max parabolic in an orthogonal group.  You just write it down!