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A proof of Razmyslov’s theorem

Notes for our seminar Lex Schrijver

Let d, n ∈ N. Let ρ be the representation Sn→ End(Cd)⊗n given by (1) ρ(π)(x1⊗ · · · ⊗ xn) = xπ(1)⊗ · · · ⊗ xπ(n)

for π ∈ Sn and x1, . . . , xn∈ Cd. Note that

(2) tr(ρ(π)(X1⊗ . . . ⊗ Xn)) = Y

ω∈Ωπ

tr(

|ω|

Y

i=1

Xπi(mω))

for π ∈ Sn and X1, . . . , Xn∈ End(Cd), where Ωπ is the set of orbits of π and where for any ω ∈ Ωπ, mω is the smallest (equivalently: an arbitrary) element of ω.

For each λ ` n, let Kλ be the isotypical component of CSn corresponding to the irre- ducible representation rλ. Define

(3) L≤d := M

λ`n height(λ)≤d

Kλ and L>d := M

λ`n height(λ)>d

Kλ.

Then (cf., e.g., [1]):

(4) Ker(ρ) = L>d.

This implies for any two subspaces U, V of CSn:

(5) ρ(U ) ⊆ ρ(V ) if and only if L>d+ U ⊆ L>d+ V .

Let An be the alternating subgroup of Sn, and let Acn := Sn\ An. Consider the linear function ϕ : CSn→ CSn determined by

(6) ϕ(π) := sgn(π)π for π ∈ Sn. So ϕ2 = id.

Lemma 1. For each λ ` n: ϕ(Kλ) = Kλ.

Proof. Let Y and Y be the Young shapes corresponding to λ and λ, respectively, let T : Y → Y be defined by T (i, j) = (j, i) for (i, j) ∈ Y , let ˜ϕ : SY → SY be defined by

˜

ϕ(π) = sgn(π)π for π ∈ SY, let HY and VY be the subgroups of SY of row-stable and column- stable permutations of Y , respectively, and let hY :=P

h∈HY h and vY :=P

v∈VY sgn(v)v.

Then

(7) ϕ(v˜ YhY) = X

v∈VY

X

h∈HY

sgn(v) ˜ϕ(vh) = X

v∈VY

X

h∈HY

sgn(v)sgn(vh)vh = X

v∈VY

X

h∈HY

sgn(h)vh = X

h∈HY ∗

X

v∈VY ∗

sgn(v)T−1hvT = T−1hYvYT.

Hence for y, z : [n] → Y , one has

1

(2)

(8) ϕ(y−1vYhYz) = sgn(y−1z)y−1ϕ(vYhY)z = sgn(y−1z)y−1T−1hYvYT z ∈ Kλ. As Kλ is spanned by all y−1vYhYz with bijections y, z : [n] → Y , we have the lemma.

Lemma 2. For each subspace U of CSn: U ∩ ϕ(U ) = (U ∩ CAn) + (U ∩ CAcn).

Proof. If x ∈ U ∩ ϕ(U ), then ϕ(x) ∈ U , hence x = 12(x + ϕ(x)) +12(x − ϕ(x)) ∈ (U ∩ CAn) + (U ∩ CAcn).

Conversely, if x ∈ U ∩CAnthen x = ϕ(x), hence x ∈ U ∩ϕ(U ). Similarly, if x ∈ U ∩CAcn

then x = −ϕ(x), hence x ∈ U ∩ ϕ(U ).

Theorem 1. ρ(CAn) = ρ(CSn) = ρ(CAcn) if and only if n > d2.

Proof. ρ(CAn) = ρ(CSn) = ρ(CAcn) ⇐⇒ L(5) >d + CAn = CSn = L>d + CAcn ⇐⇒

L≤d ∩ CAcn = {0} = L≤d ∩ CAn

Lemma 2

⇐⇒ L≤d ∩ ϕ(L≤d) = {0} Lemma 1⇐⇒ no λ ` n satisfies height(λ) ≤ d and height(λ) ≤ d ⇐⇒ n > d2.

This implies the result of Razmyslov [2]:

Corollary 1a. If n > d2, then tr(X1· · · Xn) is a linear combination of products of traces of products of fewer than n of the Xi, where X1, . . . , Xn are variable d × d matrices.

Note that in fact if n > d2, then tr(X1· · · Xn) is a linear combination of products of an even number of traces of products of the Xi.

By the above, the conclusion of Corollary 1a holds for some fixed n and d if and only if L>d+ U = CSn, where U is the subspace of CSn spanned by the permutations with at least two orbits. Note that L>d+ U = CSn is equivalent to L≤d∩ V = {0}, where V is the subspace of CSn spanned by the permutations with only one orbit.

References

[1] P. Cvitanovi´c, Group Theory: Birdtracks, Lie’s, and Exceptional Groups, Princeton University Press, Princeton, 2008.

[2] Ju.P. Razmyslov, Trace identities of full matrix algebras over a field of characteristic zero [in Russian], Izvestiya Akademii Nauk SSSR. Seriya Matematicheskaya 38 (1974) 723–756 [English translation: Mathematics of the USSR. Izvestija 8 (1974) 727–760].

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