Starting point: Range preservers

Range-compatible homomorphisms on matrix spaces

Clément de Seguins Pazzis

Preservers everywhere conference, Szeged, 2017

Starting point: Range preservers

Let

f : M n,p ( F ) → M n,p ( F ) be a linear range preserver i.e.

∀ M, Im f (M) = Im M.

Then, for some Q ∈ GL p ( F ).

f : M 7→ MQ.

Starting point: Range preservers

Let

f : M n,p ( F ) → M n,p ( F ) be a linear range preserver i.e.

∀ M, Im f (M) = Im M.

Then, for some Q ∈ GL p ( F ).

f : M 7→ MQ.

Starting point: Range preservers

Let

f : M n,p ( F ) → M n,p ( F ) be a linear range preserver i.e.

∀ M, Im f (M) = Im M.

Then, for some Q ∈ GL p ( F ).

f : M 7→ MQ.

A generalization

More generally, let M linear subspace of M n,p ( F ) (say, with codim M ”small"). Which (semi-)linear maps

f : M → M n,p ( F ) are range preservers?

Note: any such map is injective.

A generalization

More generally, let M linear subspace of M n,p ( F ) (say, with codim M ”small"). Which (semi-)linear maps

f : M → M n,p ( F ) are range preservers?

Note: any such map is injective.

An approach: range restricters

Determine the (semi)-linear maps f : M → M n,p ( F )

that are range restricters i.e.

∀ M ∈ M , Im f (M) ⊆ Im M.

Key point: if f linear and injective, f range preserver iff f and f ⁻¹

range restricters.

An approach: range restricters

Determine the (semi)-linear maps f : M → M n,p ( F )

that are range restricters i.e.

∀ M ∈ M , Im f (M) ⊆ Im M.

Key point: if f linear and injective, f range preserver iff f and f ⁻¹

range restricters.

From range restricters to range-compatibility

Let f : M → M n,p ( F ). Then,

∀ M, f (M) =

f ₁ (M) · · · f _p (M) where f _k : M → M ^n,1 ( F ) = F ⁿ .

Then, f range restricter iff each f _k is range-compatible, i.e.

∀ M ∈ M , f _k (M) ∈ Im M.

From range restricters to range-compatibility

Let f : M → M n,p ( F ). Then,

∀ M, f (M) =

f ₁ (M) · · · f _p (M) where f _k : M → M ^n,1 ( F ) = F ⁿ .

Then, f range restricter iff each f _k is range-compatible, i.e.

∀ M ∈ M , f _k (M) ∈ Im M.

The problem

What are the range-compatible additive/semi-linear/linear maps f : M → F ⁿ ?

Obvious solutions: the local maps M 7→ MX for fixed X ∈ F ^p .

Remark: the RC homomorphisms constitute a linear subspace

The problem

What are the range-compatible additive/semi-linear/linear maps f : M → F ⁿ ?

Obvious solutions: the local maps M 7→ MX for fixed X ∈ F ^p .

Remark: the RC homomorphisms constitute a linear subspace

The problem

What are the range-compatible additive/semi-linear/linear maps f : M → F ⁿ ?

Obvious solutions: the local maps M 7→ MX for fixed X ∈ F ^p .

Remark: the RC homomorphisms constitute a linear subspace

In terms of operator spaces

Let S linear subspace of L (U , V ).

F : S → V

is range-compatible iff

∀ s ∈ S , F (s) ∈ Im s.

It is local iff

F : s 7→ s(x )

In terms of operator spaces

Let S linear subspace of L (U , V ).

F : S → V

is range-compatible iff

∀ s ∈ S , F (s) ∈ Im s.

It is local iff

F : s 7→ s(x )

Basic question

Is every range-compatible additive/semi-linear/linear map F : S → V local?

S is called RC-complete (respectively, RCL-complete) if every

RC homomorphism (resp. RC linear map) on S is local.

Basic question

Is every range-compatible additive/semi-linear/linear map F : S → V local?

S is called RC-complete (respectively, RCL-complete) if every

RC homomorphism (resp. RC linear map) on S is local.

Connexions

The problem of detecting RC-complete spaces is connected to:

range-preservers;

the structure of large spaces of matrices with bounded rank;

algebraic reflexivity.

Connexions

The problem of detecting RC-complete spaces is connected to:

range-preservers;

the structure of large spaces of matrices with bounded rank;

algebraic reflexivity.

Connexions

The problem of detecting RC-complete spaces is connected to:

range-preservers;

the structure of large spaces of matrices with bounded rank;

algebraic reflexivity.

Connexions

The problem of detecting RC-complete spaces is connected to:

range-preservers;

the structure of large spaces of matrices with bounded rank;

algebraic reflexivity.

The Basic Theorem

Theorem (Various independent authors?) The space M _n,p ( F ) is RCL-complete.

If n ≥ 2 it is also RC-complete.

Special case n = 1:

every additive map on M _n,p ( F ) is range-compatible;

a map on M _1,p ( F ) is local iff it is linear.

The Basic Theorem

Theorem (Various independent authors?) The space M _n,p ( F ) is RCL-complete.

If n ≥ 2 it is also RC-complete.

Special case n = 1:

every additive map on M _n,p ( F ) is range-compatible;

a map on M _1,p ( F ) is local iff it is linear.

RC maps on spaces of nonsingular matrices

Set

M =

a − b b a

| (a, b) ∈ R ²

≃ C . Then,

F :

a − b b a

7→

a 0

linear, range-compatible, non-local.

RC maps on spaces of nonsingular matrices

Set

M =

a − b b a

| (a, b) ∈ R ²

≃ C . Then,

F :

a − b b a

7→

a 0

linear, range-compatible, non-local.

More generally, assume that n = p, dim M ≥ 2 and M ⊂ { 0 } ∪ GL n ( F ).

Then:

Every linear map F : M → F ⁿ is local.

The space of all local maps on M has dimension

≤ n < dim L ( M , F ⁿ ).

More generally, assume that n = p, dim M ≥ 2 and M ⊂ { 0 } ∪ GL n ( F ).

Then:

Every linear map F : M → F ⁿ is local.

The space of all local maps on M has dimension

≤ n < dim L ( M , F ⁿ ).

More generally, assume that n = p, dim M ≥ 2 and M ⊂ { 0 } ∪ GL n ( F ).

Then:

Every linear map F : M → F ⁿ is local.

The space of all local maps on M has dimension

≤ n < dim L ( M , F ⁿ ).

A space of 2 × 2 triangular matrices

Set

M :=

a b 0 a

| (a, b) ∈ F ²

.

Then,

F : a b

0 a

7→

0 a

linear, range-compatible, non-local.

Symmetric matrices over F ₂

Set M = S _n ( F ). For

S = (s _i,j ) _1≤i,j≤n ∈ M , set

∆(S) =









 s _1,1 s _2,2 .. . s n,n











(diagonal vector).

If F = F , then ∆ linear, non-local (if n > 1), range-compatible.

Symmetric matrices over F ₂

Set M = S _n ( F ). For

S = (s _i,j ) _1≤i,j≤n ∈ M , set

∆(S) =









 s _1,1 s _2,2 .. . s n,n











(diagonal vector).

If F = F , then ∆ linear, non-local (if n > 1), range-compatible.

Why is ∆ RC?

Let X ∈ Ker S. Prove that ∆(S) ⊥ X . Compute 0 = X ^T SX

=

n

X

i=1

s _i,i x _i ²

=

n

X

i=1

s ² _i,i x _i ²

= n

X

i=1

s _i,i x _i

2

Why is ∆ RC?

Let X ∈ Ker S. Prove that ∆(S) ⊥ X . Compute 0 = X ^T SX

=

n

X

i=1

s _i,i x _i ²

=

n

X

i=1

s ² _i,i x _i ²

= n

X

i=1

s _i,i x _i

2

Why is ∆ RC?

Let X ∈ Ker S. Prove that ∆(S) ⊥ X . Compute 0 = X ^T SX

=

n

X

i=1

s _i,i x _i ²

=

n

X

i=1

s ² _i,i x _i ²

= n

X

i=1

s _i,i x _i

2

Why is ∆ RC?

Let X ∈ Ker S. Prove that ∆(S) ⊥ X . Compute 0 = X ^T SX

=

n

X

i=1

s _i,i x _i ²

=

n

X

i=1

s ² _i,i x _i ²

= n

X

i=1

s _i,i x _i

2

Why is ∆ RC?

Let X ∈ Ker S. Prove that ∆(S) ⊥ X . Compute 0 = X ^T SX

=

n

X

i=1

s _i,i x _i ²

=

n

X

i=1

s ² _i,i x _i ²

= n

X

i=1

s _i,i x _i

2

Why is ∆ RC?

Let X ∈ Ker S. Prove that ∆(S) ⊥ X . Compute 0 = X ^T SX

=

n

X

i=1

s _i,i x _i ²

=

n

X

i=1

s ² _i,i x _i ²

= n

X

i=1

s _i,i x _i

2

Generalization to perfect fields with characteristic 2

If F perfect field with char( F ) = 2,

S ∈ S n ( F ) 7→







√ s _1,1 .. .

√ s _n,n





 = p

∆(S)

is semi-linear and RC.

Generalization to fields with characteristic 2

Further generalization: with char( F ) = 2, let f : F → F be root-linear i.e.

∀ (α, t) ∈ F ² , f (α ² t) = α f (t).

Then,

S ∈ S n ( F ) 7→ ∆(S) ^f :=









 f (s _1,1 ) f (s 2,2 )

.. . f(s n,n )











is additive and RC (not obvious!).

Generalization to fields with characteristic 2

Further generalization: with char( F ) = 2, let f : F → F be root-linear i.e.

∀ (α, t) ∈ F ² , f (α ² t) = α f (t).

Then,

S ∈ S n ( F ) 7→ ∆(S) ^f :=









 f (s _1,1 ) f (s 2,2 )

.. . f(s n,n )











is additive and RC (not obvious!).

First classification theorem

Theorem (First classification theorem) Let M ⊂ M ^n,p ( F ) with codim M ≤ n − 2.

Then, M is RC-complete.

for linear maps:

C. de Seguins Pazzis, The classification of large spaces of matrices with bounded rank, Israel J. Math. 208 (2015) 219–259.

for homomorphisms:

C. de Seguins Pazzis, Range-compatible homomorphisms on matrix spaces,

First classification theorem

Theorem (First classification theorem) Let M ⊂ M ^n,p ( F ) with codim M ≤ n − 2.

Then, M is RC-complete.

for linear maps:

C. de Seguins Pazzis, The classification of large spaces of matrices with bounded rank, Israel J. Math. 208 (2015) 219–259.

for homomorphisms:

C. de Seguins Pazzis, Range-compatible homomorphisms on matrix spaces,

First classification theorem

Theorem (First classification theorem) Let M ⊂ M ^n,p ( F ) with codim M ≤ n − 2.

Then, M is RC-complete.

for linear maps:

C. de Seguins Pazzis, The classification of large spaces of matrices with bounded rank, Israel J. Math. 208 (2015) 219–259.

for homomorphisms:

C. de Seguins Pazzis, Range-compatible homomorphisms on matrix spaces,

Why is the bound n − 2 optimal?

→ Case n = 1 there are non-linear homomorphisms M 1,p ( F ) → F iff F nonprime.

Generalization: assume ϕ : F → F nonlinear homomorphism.

Then F :

x [?] 1×p−1

[0] _(n−1)×1 [?] (n−1)×(p−1)

7→

ϕ(x ) [0] _(n−1)×1

.

is a non-linear RC homomorphism.

Why is the bound n − 2 optimal?

→ Case n = 1 there are non-linear homomorphisms M 1,p ( F ) → F iff F nonprime.

Generalization: assume ϕ : F → F nonlinear homomorphism.

Then F :

x [?] 1×p−1

[0] _(n−1)×1 [?] (n−1)×(p−1)

7→

ϕ(x ) [0] _(n−1)×1

.

is a non-linear RC homomorphism.

The Second Classification Theorem

Bound n − 2 non-optimal for linear maps!

Theorem (Second classification theorem)

Assume | F | > 2. Let M ⊂ M ^n,p ( F ) with codim M ≤ 2n − 3.

Then M is RCL-complete.

For | F | = 2, the proper upper bound is 2n − 4 (counter-example

connected to the RC maps on S ₂ ( F ₂ )).

The Second Classification Theorem

Bound n − 2 non-optimal for linear maps!

Theorem (Second classification theorem)

Assume | F | > 2. Let M ⊂ M ^n,p ( F ) with codim M ≤ 2n − 3.

Then M is RCL-complete.

For | F | = 2, the proper upper bound is 2n − 4 (counter-example

connected to the RC maps on S ₂ ( F ₂ )).

The Second Classification Theorem

Bound n − 2 non-optimal for linear maps!

Theorem (Second classification theorem)

Assume | F | > 2. Let M ⊂ M ^n,p ( F ) with codim M ≤ 2n − 3.

Then M is RCL-complete.

For | F | = 2, the proper upper bound is 2n − 4 (counter-example

connected to the RC maps on S ₂ ( F ₂ )).

Optimality of the 2n − 3 bound

Assume that p ≥ 2. Take

F :





a b [?] _1×(p−2)

0 a [?] _1×(p−2)

[0] _(n−2)×1 [0] _(n−2)×1 [?] (n−2)×(p−2)



 7→



 0 a [0] _(n−2)×1



 .

Then F is linear, RC, non-local.

Split spaces

Let M ¹ ⊂ M n,p ( F ) and M ² ⊂ M n,q ( F ). Set

M 1

a M 2 =

M ₁ M ₂

| (M ₁ , M ₂ ) ∈ M 1 × M 2

. Let F : M 1 `

M 2 → F ⁿ additive. Then, F :

M 1 M 2

7→ F ₁ (M ₁ ) + F ₂ (M ₂ )

with F ₁ : M ¹ → F ⁿ and F ₂ : M ² → F ⁿ additive. We write F = F a

F .

Split spaces

Let M ¹ ⊂ M n,p ( F ) and M ² ⊂ M n,q ( F ). Set

M 1

a M 2 =

M ₁ M ₂

| (M ₁ , M ₂ ) ∈ M 1 × M 2

. Let F : M 1 `

M 2 → F ⁿ additive. Then, F :

M 1 M 2

7→ F ₁ (M ₁ ) + F ₂ (M ₂ )

with F ₁ : M ¹ → F ⁿ and F ₂ : M ² → F ⁿ additive. We write F = F a

F .

Moreover:

F ₁ `

F ₂ linear ⇔ F ₁ and F ₂ linear;

F ₁ `

F ₂ RC ⇔ F ₁ and F ₂ RC;

F ₁ `

F ₂ local ⇔ F ₁ and F ₂ local.

Bottom line:

M RC-complete iff M 1 and M 2 RC-complete.

M RCL-complete iff M ¹ and M ² RCL-complete.

Moreover:

F ₁ `

F ₂ linear ⇔ F ₁ and F ₂ linear;

F ₁ `

F ₂ RC ⇔ F ₁ and F ₂ RC;

F ₁ `

F ₂ local ⇔ F ₁ and F ₂ local.

Bottom line:

M RC-complete iff M 1 and M 2 RC-complete.

M RCL-complete iff M ¹ and M ² RCL-complete.

Proof of the basic theorem

We have M ^n,p ( F ) = F ⁿ ` F <sup>n</sup> `

· · · `

F ⁿ (with p summands). It suffices to consider the case p = 1!!

Lemma

Any linear subspace M ⊂ F ⁿ with dim M ≥ 2 is RC-complete!

Proof.

Let F : M → F ⁿ RC homomorphism. Then,

∀ X ∈ M , ∃ λ X ∈ F : F (X ) = λ X X .

Classically, F : X 7→ λX for some fixed λ ∈ F , i.e.

Proof of the basic theorem

We have M ^n,p ( F ) = F ⁿ ` F <sup>n</sup> `

· · · `

F ⁿ (with p summands). It suffices to consider the case p = 1!!

Lemma

Any linear subspace M ⊂ F ⁿ with dim M ≥ 2 is RC-complete!

Proof.

Let F : M → F ⁿ RC homomorphism. Then,

∀ X ∈ M , ∃ λ X ∈ F : F (X ) = λ X X .

Classically, F : X 7→ λX for some fixed λ ∈ F , i.e.

Proof of the basic theorem

We have M ^n,p ( F ) = F ⁿ ` F <sup>n</sup> `

· · · `

F ⁿ (with p summands). It suffices to consider the case p = 1!!

Lemma

Any linear subspace M ⊂ F ⁿ with dim M ≥ 2 is RC-complete!

Proof.

Let F : M → F ⁿ RC homomorphism. Then,

∀ X ∈ M , ∃ λ X ∈ F : F (X ) = λ X X .

Classically, F : X 7→ λX for some fixed λ ∈ F , i.e.

Quotient space technique

Basic idea: induction on the number p of rows. Delete a row (projection). Let F : M → F ⁿ RC homomorphism.

M =

K (M) [?] 1×p

7→ F (M) =

C(M)

?

. Then C(M) ∈ Im K (M).

Moreover, C (M) additive function of K (M)!

Indeed, K (M) = 0 ⇒ C(M) = 0.

Thus,

F (M) =

G(K (M))

?

.

Quotient space technique

Basic idea: induction on the number p of rows. Delete a row (projection). Let F : M → F ⁿ RC homomorphism.

M =

K (M) [?] 1×p

7→ F (M) =

C(M)

?

. Then C(M) ∈ Im K (M).

Moreover, C (M) additive function of K (M)!

Indeed, K (M) = 0 ⇒ C(M) = 0.

Thus,

F (M) =

G(K (M))

?

.

Quotient space technique

Basic idea: induction on the number p of rows. Delete a row (projection). Let F : M → F ⁿ RC homomorphism.

M =

K (M) [?] 1×p

7→ F (M) =

C(M)

?

. Then C(M) ∈ Im K (M).

Moreover, C (M) additive function of K (M)!

Indeed, K (M) = 0 ⇒ C(M) = 0.

Thus,

F (M) =

G(K (M))

?

.

Quotient space technique

Basic idea: induction on the number p of rows. Delete a row (projection). Let F : M → F ⁿ RC homomorphism.

M =

K (M) [?] 1×p

7→ F (M) =

C(M)

?

. Then C(M) ∈ Im K (M).

Moreover, C (M) additive function of K (M)!

Indeed, K (M) = 0 ⇒ C(M) = 0.

Thus,

F (M) =

G(K (M))

?

.

Quotient space technique

Basic idea: induction on the number p of rows. Delete a row (projection). Let F : M → F ⁿ RC homomorphism.

M =

K (M) [?] 1×p

7→ F (M) =

C(M)

?

. Then C(M) ∈ Im K (M).

Moreover, C (M) additive function of K (M)!

Indeed, K (M) = 0 ⇒ C(M) = 0.

Thus,

F (M) =

G(K (M))

?

.

Quotient space technique

Basic idea: induction on the number p of rows. Delete a row (projection). Let F : M → F ⁿ RC homomorphism.

M =

K (M) [?] 1×p

7→ F (M) =

C(M)

?

. Then C(M) ∈ Im K (M).

Moreover, C (M) additive function of K (M)!

Indeed, K (M) = 0 ⇒ C(M) = 0.

Thus,

F (M) =

G(K (M))

?

.

In terms of operators

Choose x ∈ V \ { 0 } , set D := F x (line spanned by x ).

For s ∈ L (U , V ), set

s mod D : x 7→ s(x ) ∈ V /D.

Then, S mod D is a linear subspace of L (U, V /D), and any RC additive F : S → V induces

F mod D : S mod D → V /D so that the diagram commutes

S ^{F / /} V

In terms of operators

Choose x ∈ V \ { 0 } , set D := F x (line spanned by x ).

For s ∈ L (U , V ), set

s mod D : x 7→ s(x ) ∈ V /D.

Then, S mod D is a linear subspace of L (U, V /D), and any RC additive F : S → V induces

F mod D : S mod D → V /D so that the diagram commutes

S ^{F / /} V

In terms of operators

Choose x ∈ V \ { 0 } , set D := F x (line spanned by x ).

For s ∈ L (U , V ), set

s mod D : x 7→ s(x ) ∈ V /D.

Then, S mod D is a linear subspace of L (U, V /D), and any RC additive F : S → V induces

F mod D : S mod D → V /D so that the diagram commutes

S ^{F / /} V

Idea 1: F is known if F mod D ₁ and F mod D ₂ are known for distinct lines D ₁ and D ₂ .

Idea 2: codim( S mod D) ≤ codim( S ) (rank theorem). Equality holds iff

∀ s ∈ L (U, V ), Im s ⊂ D ⇒ s ∈ S .

When, codim( S mod D) < codim( S ), say that D is adapted to S .

Idea 1: F is known if F mod D ₁ and F mod D ₂ are known for distinct lines D ₁ and D ₂ .

Idea 2: codim( S mod D) ≤ codim( S ) (rank theorem). Equality holds iff

∀ s ∈ L (U, V ), Im s ⊂ D ⇒ s ∈ S .

When, codim( S mod D) < codim( S ), say that D is adapted to S .

Application to the first classification theorem

Inductive proof. Theorem is trivial if n < 2.

Assume now that n ≥ 2. Let S ⊂ L (U, V ) with codim S ≤ dim V − 2.

Assume that there are distinct lines D ₁ = F y ₁ and D ₂ = F y ₂ adapted to S . Let F : S → V a RC homomorphism.

Then, F mod D ₁ and F mod D ₂ are local (by induction).

Yields x ₁ , x ₂ ∈ U s.t. for all s ∈ S ,

F (s) = s(x ₁ ) mod D ₁ and F (s) = s(x ₂ ) mod D _2.

Application to the first classification theorem

Inductive proof. Theorem is trivial if n < 2.

Assume now that n ≥ 2. Let S ⊂ L (U, V ) with codim S ≤ dim V − 2.

Assume that there are distinct lines D ₁ = F y ₁ and D ₂ = F y ₂ adapted to S . Let F : S → V a RC homomorphism.

Then, F mod D ₁ and F mod D ₂ are local (by induction).

Yields x ₁ , x ₂ ∈ U s.t. for all s ∈ S ,

F (s) = s(x ₁ ) mod D ₁ and F (s) = s(x ₂ ) mod D _2.

Application to the first classification theorem

Inductive proof. Theorem is trivial if n < 2.

Assume now that n ≥ 2. Let S ⊂ L (U, V ) with codim S ≤ dim V − 2.

Assume that there are distinct lines D ₁ = F y ₁ and D ₂ = F y ₂ adapted to S . Let F : S → V a RC homomorphism.

Then, F mod D ₁ and F mod D ₂ are local (by induction).

Yields x ₁ , x ₂ ∈ U s.t. for all s ∈ S ,

F (s) = s(x ₁ ) mod D ₁ and F (s) = s(x ₂ ) mod D _2.

Application to the first classification theorem

Inductive proof. Theorem is trivial if n < 2.

Assume now that n ≥ 2. Let S ⊂ L (U, V ) with codim S ≤ dim V − 2.

Assume that there are distinct lines D ₁ = F y ₁ and D ₂ = F y ₂ adapted to S . Let F : S → V a RC homomorphism.

Then, F mod D ₁ and F mod D ₂ are local (by induction).

Yields x ₁ , x ₂ ∈ U s.t. for all s ∈ S ,

F (s) = s(x ₁ ) mod D ₁ and F (s) = s(x ₂ ) mod D _2.

Application to the first classification theorem

Inductive proof. Theorem is trivial if n < 2.

Assume now that n ≥ 2. Let S ⊂ L (U, V ) with codim S ≤ dim V − 2.

Assume that there are distinct lines D ₁ = F y ₁ and D ₂ = F y ₂ adapted to S . Let F : S → V a RC homomorphism.

Then, F mod D ₁ and F mod D ₂ are local (by induction).

Yields x ₁ , x ₂ ∈ U s.t. for all s ∈ S ,

F (s) = s(x ₁ ) mod D ₁ and F (s) = s(x ₂ ) mod D _2.

First classification theorem, continued

If x ₁ = x ₂ then F : s 7→ s(x 1 ) QED.

Assume x ₁ 6 = x ₂ . Then, for all s ∈ S ,

F (s) = s(x ₁ ) mod D ₁ + D ₂ and F (s) = s(x ₂ ) mod D ₁ + D ₂ whence

∀ s ∈ S , s(x 1 − s ₂ ) ∈ D ₁ + D ₂ = Span(y 1 , y ₂ ).

First classification theorem, continued

If x ₁ = x ₂ then F : s 7→ s(x 1 ) QED.

Assume x ₁ 6 = x ₂ . Then, for all s ∈ S ,

F (s) = s(x ₁ ) mod D ₁ + D ₂ and F (s) = s(x ₂ ) mod D ₁ + D ₂ whence

∀ s ∈ S , s(x 1 − s ₂ ) ∈ D ₁ + D ₂ = Span(y 1 , y ₂ ).

First classification theorem, continued

Extend x ₁ − x ₂ into a basis B = (x 1 − x ₂ , e ₂ , . . . , e _p ) of U, (y ₁ , y ₂ ) into a basis C = (y ₁ , y ₂ , . . . , y n ) of V .

In those bases

S ←→

( 



? [?] _1×(p−1)

? [?] _2×(p−1) [0] _(n−2)×1 [?] (n−2)×(p−1)



 )

=: M .

Now,

M = ( F ² × { 0 } ) a F ⁿ a

· · · a

F ⁿ

First classification theorem, continued

Extend x ₁ − x ₂ into a basis B = (x 1 − x ₂ , e ₂ , . . . , e _p ) of U, (y ₁ , y ₂ ) into a basis C = (y ₁ , y ₂ , . . . , y n ) of V .

In those bases

S ←→

( 



? [?] _1×(p−1)

? [?] _2×(p−1) [0] _(n−2)×1 [?] (n−2)×(p−1)



 )

=: M .

Now,

M = ( F ² × { 0 } ) a F ⁿ a

· · · a

F ⁿ

First classification theorem: The existence of two adapted lines

Suppose that there is no more than 1 adapted line.

Then, for some basis (y ₁ , . . . , y _n ) ∈ V ⁿ s.t.

∀ i, { s ∈ L (U, V ) : Im s ⊂ F y _i } ⊂ S . By summing,

S = L (U, V )

and S is RC-complete. QED.

First classification theorem: The existence of two adapted lines

Suppose that there is no more than 1 adapted line.

Then, for some basis (y ₁ , . . . , y _n ) ∈ V ⁿ s.t.

∀ i, { s ∈ L (U, V ) : Im s ⊂ F y _i } ⊂ S . By summing,

S = L (U, V )

and S is RC-complete. QED.

Some open problems

What happens beyond the critical bound 2n − 3?

→ probably too difficult!

Structured case: classify RC homomorphisms on subspaces of S n ( F ) and A n ( F ).

Theorem

Let M ⊂ S n ( F ) with char( F ) 6 = 2. If codim M ≤ n − 2 then M is RC-complete.

Let M ⊂ A n ( F ). If codim M ≤ n − 3 then M is RC-complete.

Optimal bounds!

Open problem: what is the least k such that every subspace

Some open problems

What happens beyond the critical bound 2n − 3?

→ probably too difficult!

Structured case: classify RC homomorphisms on subspaces of S n ( F ) and A n ( F ).

Theorem

Let M ⊂ S n ( F ) with char( F ) 6 = 2. If codim M ≤ n − 2 then M is RC-complete.

Let M ⊂ A n ( F ). If codim M ≤ n − 3 then M is RC-complete.

Optimal bounds!

Open problem: what is the least k such that every subspace

Some open problems

What happens beyond the critical bound 2n − 3?

→ probably too difficult!

Structured case: classify RC homomorphisms on subspaces of S n ( F ) and A n ( F ).

Theorem

Let M ⊂ S n ( F ) with char( F ) 6 = 2. If codim M ≤ n − 2 then M is RC-complete.

Let M ⊂ A n ( F ). If codim M ≤ n − 3 then M is RC-complete.

Optimal bounds!

Open problem: what is the least k such that every subspace

Some open problems

What happens beyond the critical bound 2n − 3?

→ probably too difficult!

Structured case: classify RC homomorphisms on subspaces of S n ( F ) and A n ( F ).

Theorem

Let M ⊂ S n ( F ) with char( F ) 6 = 2. If codim M ≤ n − 2 then M is RC-complete.

Let M ⊂ A n ( F ). If codim M ≤ n − 3 then M is RC-complete.

Optimal bounds!

Open problem: what is the least k such that every subspace