1. PIT Questions in Invariant Theory
Rafael Oliveira
University of Toronto
PIT Questions in Invariant Theory

2. 1 Outline Geom. Inv. Theory Inv. Theory and PIT Examples & Results
Future Work

3. Invariant Theory - Basics
Let 𝑽 be a vector space and 𝑮 be a group acting (linearly) on it
𝑉= ℂ 𝑛 and 𝐺= 𝑆 𝑛 permutes coordinates
𝑉=𝑀𝑎 𝑡 𝑛 (ℂ) and 𝐺=𝑆𝐿 𝑛 ×𝑆𝐿(𝑛) acts mult. on left-right
An invariant polynomial is a polynomial which doesn’t change by action of 𝐺
𝑝 𝑔⋅𝑥 =𝑝 𝑥 , ∀ 𝑔∈𝐺
𝑉= ℂ 𝑛 and 𝐺= 𝑆 𝑛 invariants gen. by symmetric polynomials
𝑉=𝑀𝑎 𝑡 𝑛 (ℂ) and 𝐺=𝑆𝐿 𝑛 ×𝑆𝐿(𝑛) gen. by determinant
Given 𝑣∈𝑉, its orbit is the set
𝒪 𝑣 ≝{𝑔⋅𝑣∣𝑔∈𝐺}
And the closure 𝒪 𝑣 is the orbit with its limit points.
(Also, zero set of all polynomials vanishing on G⋅𝑣).

4. Basic Questions
Given vector space 𝑽and group 𝑮 acting (linearly) on it
Compute generators to invariant polynomials
Find relations among generators (a.k.a. syzygies)
Many more
Orbit Closure Intersection (OCI): given 𝑣, 𝑤∈𝑉, is 𝒪 𝑣 ∩ 𝒪 𝑤 =∅?
Thm [Mum’65]: orbit closures don’t intersect iff
∃ 𝑝∈ℂ 𝑉 𝐺 s.t. 𝑝 𝑣 ≠𝑝(𝑤)
Suffices to eval. on random linear comb. of gen. of ℂ 𝑉 𝐺
How can we deterministically find such a polynomial? Is this the only way to solve it?
Null-cone Problem: given 𝑣∈𝑉, is 0∈ 𝒪 𝑣 ?
How do these problems relate to PIT?

5. Invariant Theory and PIT
Input description:
Group 𝐆, vector space 𝐕 given by natural encoding, 1 dim 𝑉 +dim⁡(𝐺)
Example: 𝑮=𝑮𝑳 𝒏 , 𝑽=𝑴𝒂 𝒕 𝒏 (ℂ) and action 𝒈⋅𝑨=𝒈𝑨 𝒈 −𝟏
Problems of interest:
Null-cone problem
OCI
Efficiently (and succinctly) computing generators of ℂ 𝑽 𝑮
Orbit Closure Containment (OCC)
More generally we could talk about “explicit varieties” but then we won’t necessarily have invariant theory involved. So probably not good for this talk. Figure out a better theme for such a talk.
Can we get a hierarchy of varieties? “Succinct” < “explicit” < …
Remarks:
Null-cone problem and OCI witnessed by inv. Polynomials
OCC seems much harder (Markus’ talk)
Examples of OCC problems
𝑽𝑷 𝒗𝒔 𝑽𝑵𝑷
Border Rank

6. Invariant Theory and PIT
Basic structural questions (and results):
Is ℂ 𝑽 𝑮 finitely generated?
Thm [Hil’90, 93]: Yes!
Upper bounds on degree of generators?
Thm [Popov’81, Derksen’01]: exponential degree bounds!
General construction of invariants?
Reynolds operator (Cayley Omega, Casimir)
Previous algorithms doubly exponential or exponential until Ketan came along and gave the right complexity theoretic notion of succinctness
Remarks:
Reynolds operator general, but not “efficient”. For algs. need “simple invariants”
Hard to prove better degree bounds even for special cases
Previous algs for Null Cone and OCI were doubly exponential

7. Invariant Theory and PIT
Succinct generators [Mul’12]: ℂ 𝑽 𝑮 has 𝓒-succinct generators, if there is a small circuit 𝑭 𝒗, 𝒚 ∈𝓒 such that
𝑭 𝒗, 𝒚 = 𝒊 𝒇 𝒊 𝒗 𝒈 𝒊 (𝒚)
𝒇 𝒊 𝒗 𝒊 is a set of generators of ℂ 𝑽 𝑮
𝒈 𝒊 𝒚 𝒊 are linearly independent
If ℂ 𝑽 𝑮 has 𝓒-succinct generators, then Null Cone and OCI can be solved by solving PIT for 𝓒!
Circuit class C is a “natural circuit class” (i.e.) closed under restrictions, and for PIT purposes of polynomial degree.
Note that for nullcone and OCI the polynomial is a polynomial only in the auxiliary variables y!
Remark: enough to take C-succinct separators (where the f_i’s are separating invariants)
To test if 𝒖 in null cone: test if 𝑭 𝒖, 𝒚 ≡𝟎
To test OCI, that is, 𝓞 𝒖 ∩ 𝓞 𝒘 ≠∅: test if 𝑭 𝒖, 𝒚 −𝑭 𝒘, 𝒚 ≡𝟎
Remark: analytic approach may sometimes not need degree bounds on the circuit class 𝒞 for certain problems.
[GGOW’16] solution to the null cone

8. Simultaneous Conjugation (SC)
Setup: group 𝐺=𝑆𝐿(𝑛), vector space 𝑉=𝑀𝑎 𝑡 𝑛 ℂ 𝑚
𝑔⋅ 𝐴 1 ,…, 𝐴 𝑚 →(𝑔 𝐴 1 𝑔 −1 ,…,𝑔 𝐴 𝑚 𝑔 −1 )
Null Cone: 𝐴 1 ,…, 𝐴 𝑚 ∈𝑉 with rational entries
OCI: 𝐴 1 ,…, 𝐴 𝑚 , ( 𝐵 1 ,…, 𝐵 𝑚 )∈𝑉 with rational entries
Invariants [Pro’76, Raz’74, For’86]: generated by
𝑡𝑟 𝑋 𝑖 1 ⋯ 𝑋 𝑖 ℓ , where 1≤ℓ≤ 𝑛 2 , 𝑖 𝑗 ∈[𝑚]
Have polynomial degree bounds!
Is the set above succinct?
Procesi, Razmyslov, Formanek

9. Simultaneous Conjugation (SC)
Invariants [Pro’76, Raz’74, For’86]: generated by
𝑡𝑟 𝑋 𝑖 1 ⋯ 𝑋 𝑖 ℓ , where 1≤ℓ≤ 𝑛 2 , 𝑖 𝑗 ∈[𝑚]
Succinctness [Mul’12, FS’12]:
𝑍 𝑗 = 𝑦 𝑗0 ⋅𝐼+ 𝑖∈[𝑚] 𝑦 𝑗𝑖 ⋅ 𝑋 𝑖 , 1≤𝑗≤ 𝑛 2
𝐹 𝑋 1 ,…, 𝑋 𝑚 , 𝑦 10 ,… 𝑦 𝑛 2 𝑚 = 𝑡𝑟 𝑍 1 ⋯ 𝑍 𝑛 2 =∑ ∏ 𝑦 𝑗 𝑖 𝑗 ⋅𝑡𝑟 ∏ 𝑋 𝑖 𝑗
Procesi, Razmyslov, Formanek

10. Simultaneous Conjugation (SC)
Null Cone: 𝐴 1 ,…, 𝐴 𝑚 ∈𝑉 with rational entries
OCI: 𝐴 1 ,…, 𝐴 𝑚 , ( 𝐵 1 ,…, 𝐵 𝑚 )∈𝑉 with rational entries
Succinctness:
𝐹 𝑋 1 ,…, 𝑋 𝑚 , 𝑦 10 ,… 𝑦 𝑛 2 𝑚 = 𝑡𝑟 𝑍 1 ⋯ 𝑍 𝑛 2 =∑ ∏ 𝑦 𝑗 𝑖 𝑗 ⋅𝑡𝑟 ∏ 𝑋 𝑖 𝑗
OCI: by [Mum’65] and succinctness, enough to check
𝐹 𝐴 1 ,…, 𝐴 𝑚 , 𝑦 10 ,… 𝑦 𝑛 2 𝑚 −𝐹 𝐵 1 ,…, 𝐵 𝑚 , 𝑦 10 ,… 𝑦 𝑛 2 𝑚 ≠0
Polynomial above zero iff 𝑡𝑟 𝐴 𝑖 1 ⋯ 𝐴 𝑖 ℓ =𝑡𝑟 𝐵 𝑖 1 ⋯ 𝐵 𝑖 ℓ on all gen.
[FS’12]: polynomial above is ROABP over 𝒚 variables!
[RS’05] white-box PIT in P
[FS’12] black-box PIT in quasi-P
Procesi, Razmyslov, Formanek
Perhaps mention here that FS does not provide a separating invariant in their algorithm.

11. Left-Right (LR) Action
Setup: group 𝐺=𝑆𝐿 𝑛 ×𝑆𝐿(𝑛), vector space 𝑉=𝑀𝑎 𝑡 𝑛 ℂ 𝑚
(𝑔,ℎ)⋅ 𝐴 1 ,…, 𝐴 𝑚 →(𝑔 𝐴 1 ℎ −1 ,…,𝑔 𝐴 𝑚 ℎ −1 )
Null Cone: 𝐀= 𝐴 1 ,…, 𝐴 𝑚 ∈𝑉 with rational entries
OCI: 𝐀= 𝐴 1 ,…, 𝐴 𝑚 , 𝑩=( 𝐵 1 ,…, 𝐵 𝑚 )∈𝑉 with rational entries
Invariants [DW’00, SvdB’01, DZ’01,DM’16, IQS’16]: generated by
𝑝 𝑑 𝒀 (𝒅) ,𝑿 =𝑑𝑒𝑡 𝑖∈[𝑚] 𝑌 𝑖 (𝑑) ⊗ 𝑋 𝑖 , 𝑌 𝑖 (𝑑) ∈𝑀𝑎 𝑡 𝑑 ℂ , 𝑑≤ 𝑛 5
PIT question: 𝐀≉𝐁 iff there is 𝑑≤ 𝑛 5 s.t.
𝑝 𝑑 𝑌 1 𝑑 ,…, 𝑌 1 𝑑 ,𝑨 − 𝑝 𝑑 𝑌 1 𝑑 ,…, 𝑌 1 𝑑 ,𝑩 ≠0
𝑑∈[ 𝑛 5 ] 𝑝 𝑑 𝑌 1 𝑑 ,…, 𝑌 1 𝑑 ,𝑨 − 𝑝 𝑑 𝑌 1 𝑑 ,…, 𝑌 1 𝑑 ,𝑩 ⋅ 𝑧 𝑑 ≠0
Derksen Weyman 2000
Domokos Zubkov 2001
Schoefield van der Bergh 2001

12. LR Action – Null Cone
Setup: group 𝐺=𝑆𝐿 𝑛 ×𝑆𝐿(𝑛), vector space 𝑉=𝑀𝑎 𝑡 𝑛 ℂ 𝑚
(𝑔,ℎ)⋅ 𝐴 1 ,…, 𝐴 𝑚 →(𝑔 𝐴 1 ℎ −1 ,…,𝑔 𝐴 𝑚 ℎ −1 )
Null Cone: 𝐀= 𝐴 1 ,…, 𝐴 𝑚 ∈𝑉 with rational entries
Analytic algorithm [Gur’04, GGOW’16]: decide membership in the null cone in poly-time. No need for upper bound on the degree of invariants!
Algebraic algorithm [IQS’16]: decide membership in the null-cone in poly-time. In particular, if 𝐀 not in null cone finds 𝑌 1 𝑑 ,…, 𝑌 1 𝑑 such that
𝑝 𝑑 𝑌 1 𝑑 ,…, 𝑌 1 𝑑 ,𝑨 ≠0
In the IQS: need to put the d \in {n-1, n} condition somewhere

13. LR Action – OCI via reduction [DM’18]
Setup: group 𝐺=𝑆𝐿 𝑛 ×𝑆𝐿(𝑛), vector space 𝑉=𝑀𝑎 𝑡 𝑛 ℂ 𝑚
(𝑔,ℎ)⋅ 𝐴 1 ,…, 𝐴 𝑚 →(𝑔 𝐴 1 ℎ −1 ,…,𝑔 𝐴 𝑚 ℎ −1 )
OCI: 𝐴 1 ,…, 𝐴 𝑚 , ( 𝐵 1 ,…, 𝐵 𝑚 )∈𝑉 with rational entries
We know OCI for SC. Could we reduce OCI for LR-action to SC?
Reduction [DM’18]: OCI for LR-action is equivalent to OCI for SC under poly-time reductions.
Easy direction: OCI for SC ≤ 𝑝 OCI for LR-action.
Pf: 𝐴 1 ,…, 𝐴 𝑚 ≈ 𝑆𝐶 ( 𝐵 1 ,…, 𝐵 𝑚 ) iff 𝐼, 𝐴 1 ,…, 𝐴 𝑚 ≈ 𝐿𝑅 (𝐼, 𝐵 1 ,…, 𝐵 𝑚 )
Derksen Makam 2018
Hard direction (some easy facts):
can assume that det 𝐴 𝑖 = det 𝐵 𝑖 ∀𝑖∈[𝑚].
OCI problem doesn’t change if replace input with another elt. from same orbit

14. LR Action – Reduction to SC [DM’18]
Harder direction: OCI for LR-action ≤ 𝑝 OCI for SC.
Easy case: 𝐴 1 ,…, 𝐴 𝑚 in the null cone.
In this case OCI is equivalent to Null Cone problem.
By [GGOW’16, IQS’16] solve this in P
Another easy case: det A 1 ≠0 then 𝐴 1 ,…, 𝐴 𝑚 ≈ 𝐿𝑅 𝐵 1 ,…, 𝐵 𝑚 iff
𝐼, 𝐴 1 −1 𝐴 2 ,…, 𝐴 1 −1 𝐴 𝑚 ≈ 𝐿𝑅 (𝐼, 𝐵 1 −1 𝐵 2 ,…, 𝐵 1 −1 𝐵 𝑚 )⇔ 𝐴 1 −1 𝐴 2 ,…, 𝐴 1 −1 𝐴 𝑚 ≈ 𝑆𝐶 ( 𝐵 1 −1 𝐵 2 ,…, 𝐵 1 −1 𝐵 𝑚 )
Easy to adapt to case where 𝑠𝑝𝑎𝑛( 𝐴 𝑖 ) has invertible matrix*
Derksen Makam 2018
Note that B_1 will also be invertible because we assume that det(A_i) = det(B_i) for all I
(*) And we know how to obtain this matrix
Remaining case: 𝐴 1 ,…, 𝐴 𝑚 , ( 𝐵 1 ,…, 𝐵 𝑚 ) not in null cone and don’t span an invertible matrix.
Plan: reduce to the previous case (find inv. matrix)
Need to use blow-ups

15. LR Action – Reduction to SC [DM’18]
Blow-ups: 𝐀= 𝐴 1 ,…, 𝐴 𝑚 in 𝑀𝑎 𝑡 𝑛 ℂ 𝑚 , its 𝑘 𝑡ℎ -order blow-up is
𝑨 𝒌 = 𝑨 𝒊 ⊗ 𝑬 𝒋𝒌 , 𝑖∈ 𝑚 ; 𝑗,𝑘∈[𝑘]
Lemma [DM’18]: 𝐀, 𝐁∈𝑀𝑎 𝑡 𝑛 ℂ 𝑚 with rational entries, 𝑘≥𝑛−1
𝐀 ≈ 𝑳𝑹 𝐁⇔ 𝑨 𝒌 ≈ 𝑳𝑹 𝑩 𝒌
Remaining case: 𝐀, 𝐁 not in null cone and det 𝐴 𝑖 = det 𝐵 𝑖 =0 for all 𝑖∈ 𝑚 .
𝐀 not in null cone then there is 𝑌 𝑖 (𝑛−1) ∈𝑀𝑎 𝑡 𝑛−1 ℤ s.t.
𝑑𝑒𝑡 𝑖∈[𝑚] 𝑌 𝑖 (𝑛−1) ⊗ 𝐴 𝑖 ≠0
In particular, 𝐀 𝐧−𝟏 has an invertible element
Using lemma reduce blown-up instance to SC
Derksen Weyman 2000
Domokos Zubkov 2001
Schoefield van der Bergh 2001

16. OCI for LR-action (Analytic Result) - Outline
[AGLOW’18]: decide OCI for LR-action in poly-time. Analytic alg.
Now polynomial bounds on the degree of invariants needed!
Outline:
Scaling to DS takes us to elt of min norm in closures
Let 𝐯 and 𝒘 be elts of min norm in 𝑶 𝒗 and 𝑶 𝒘
By [KN’79] 𝑶 𝒗 and 𝑶 𝒘 intersect iff 𝐯 and 𝒘 are equivalent under “rotations” (maximal compact subgroup)
Test if 𝐯 and 𝒘 are equivalent under “rotations”W
Too much to introduce in this talk
Remark: analytic approach only gives us approximate minimizers, so need to do approximation version of steps 1-4

17. ? Thank you! Open Questions
Create a complexity theory of the computational problems in invariant theory
[DM’18] showed that the OCI for SC equivalent to OCI for LR-action
Complete problems?
More applications?
Analytic methods vs algebraic methods
what are their powers and limitations?
Thank you!