1. CapAndHomalgNotebooks
  2. SymmetricAlgebraInternalToS4rep.ipynb

binder

This is an example for explicit computations in the symmetric algebra $\mathrm{S}W$ internal to the positively graded closure $\mathrm{srep}_k(G)_+$ of the skeletal tensor category $\mathrm{srep}_k(G)$ of finite dimensional $k$-linear representations of $G$, where $G := S_4$ is the symmetric group on four points and $k = \mathbb{Q}$, its minimal splitting field. The natural permutation character on four points is the sum of the irreducible characters $𝟙+ν$, where $𝟙$ denotes the trivial character. Denote the irreducible degree $2$ character by $ρ$, the sign-character by $σ$, and set $χ := \sigma ν$. The character table of $S_4$ $$ \begin{array}{c|rrrrr} & (1) & (12) & (12)(34) & (123) & (1234) \\ \hline 𝟙 & 1 & 1 & 1 & 1 & 1 \\ σ & 1 & -1 & 1 & 1 & -1 \\ ρ & 2 & \cdot & 2 & -1 & \cdot \\ ν & 3 & 1 & -1 & \cdot & -1 \\ χ & 3 & -1 & -1 & \cdot & 1 \end{array} $$

with componentwise addition and multiplication determines the structure of $\mathrm{srep}_k(S_4)$ as a skeletal semisimple $\mathbb{Q}$-linear Abelian category with bifunctor $\otimes$.

In [1]:
using CapAndHomalg

CapAndHomalg v1.0.0
Imported OSCAR's components GAP and Singular_jll
Type: ?CapAndHomalg for more information

Using the package $\mathtt{GroupRepresentationsForCAP}$ we now construct $\mathrm{srep}_k(S_4)$, even as a tensor category over $\mathbb{Q}$:

In [2]:
LoadPackage( "GroupRepresentationsForCAP" )
In [3]:
G = SymmetricGroup( 4 )
Out[3]:
GAP: Sym( [ 1 .. 4 ] )
In [4]:
srepG = RepresentationCategory( G )
Out[4]:
GAP: The representation category of SymmetricGroup( [ 1 .. 4 ] )
In [5]:
InfoOfInstalledOperationsOfCategory( srepG )

46 primitive operations were used to derive 202 operations for this category which
* IsLinearCategoryOverCommutativeRing
* IsRigidSymmetricClosedMonoidalCategory
* IsClosedMonoidalCategory
* IsAbelianCategory
In [6]:
CommutativeRingOfLinearCategory( srepG )
Out[6]:
GAP: Q
In [7]:
irr = Irr( G );
In [8]:
𝟙 = RepresentationCategoryObject( irr[5], srepG, "𝟙" )
Out[8]:
$$𝟙$$
In [9]:
σ = RepresentationCategoryObject( irr[1], srepG, "σ" )
Out[9]:
$$σ$$
In [10]:
ρ = RepresentationCategoryObject( irr[3], srepG, "ρ" )
Out[10]:
$$ρ$$
In [11]:
ν = RepresentationCategoryObject( irr[4], srepG, "ν" )
Out[11]:
$$ν$$
In [12]:
χ = RepresentationCategoryObject( irr[2], srepG, "χ" )
Out[12]:
$$χ$$

Regardless of which choices we make, the associator and braiding cannot be given by identities. For example:

We compute the tensor product $(ρ \otimes σ) \otimes ρ = σ \oplus ρ \oplus 𝟙$:

In [13]:
TensorProduct( ρ, σ, ρ )
Out[13]:
$$σ \oplus ρ \oplus 𝟙$$

The associator $α_{ρσρ}: ρ \otimes (σ \otimes ρ) = σ \oplus ρ \oplus 𝟙 \to σ \oplus ρ \oplus 𝟙 = (ρ \otimes σ) \otimes ρ$ is nontrivial:

In [14]:
α_ρσρ = AssociatorRightToLeft( ρ, σ, ρ )
Out[14]:
$$σ \oplus ρ \oplus 𝟙 \xrightarrow{\operatorname{diag}(-1,1,-1)} σ \oplus ρ \oplus 𝟙$$

We compute the tensor product $ρ \otimes ρ = σ \oplus ρ \oplus 𝟙$:

In [15]:
TensorProduct( ρ, ρ )
Out[15]:
$$σ \oplus ρ \oplus 𝟙$$

The braiding $γ_{ρρ}: ρ \otimes ρ = σ \oplus ρ \oplus 𝟙 \to σ \oplus ρ \oplus 𝟙 = ρ \otimes ρ$ is nontrivial:

In [16]:
γ_ρρ = Braiding( ρ, ρ )
Out[16]:
$$σ \oplus ρ \oplus 𝟙 \xrightarrow{\operatorname{diag}(-1,1,1)} σ \oplus ρ \oplus 𝟙$$

Using the package $\mathtt{GradedCategories}$ we now construct the positively graded closure $\mathrm{srep}_k(S_4)_+$ of $\mathrm{srep}_k(S_4)$.

In [17]:
LoadPackage( "GradedCategories" )
In [18]:
ZsrepG = PositivelyZGradedCategory( srepG )
Out[18]:
GAP: The positively graded category of The representation category of SymmetricGroup( [ 1 .. 4 ] )

For $W = χ \equiv χ^{\{1\}} \in \operatorname{srep}_k(S_4)_+$ we use the package $\mathtt{InternalModules}$ to construct the symmetric algebra of $W$ internal to $\operatorname{srep}_k(S_4)_+$:

The objects in $\mathrm{S}^i W$ can be computed using the character table of $S_4$ augmented with the power maps:

\begin{align*} \mathrm{S}W &= \underbrace{𝟙^{\{0\}}}_{\mathrm{S}^0 W} \oplus \underbrace{χ^{\{1\}}}_{\mathrm{S}^1 W} \oplus \underbrace{\left( ρ^{\{2\}} \oplus ν^{\{2\}} \oplus 𝟙^{\{2\}} \right)}_{\mathrm{S}^2 W} \oplus \underbrace{\left( σ^{\{3\}} \oplus 2 \cdot χ^{\{3\}} \oplus ν^{\{3\}} \right)}_{\mathrm{S}^3 W} \\ & \oplus \underbrace{\left( χ^{\{4\}} \oplus 2 \cdot ρ^{\{4\}} \oplus 2 \cdot ν^{\{4\}} \oplus 2 \cdot 𝟙^{\{4\}} \right)}_{\mathrm{S}^4 W} \oplus \cdots \end{align*}
In [19]:
LoadPackage( "InternalModules" )
In [20]:
W = χ
Out[20]:
$$χ$$
In [21]:
SWMod = CategoryOfLeftSModules( W )
Out[21]:
GAP: Abelian category of left modules over the internal symmetric algebra of 1*(χ) with undecidable (mathematical) equality of morphisms and uncomputable lifts and colifts
In [22]:
SW = UnderlyingActingObject( SWMod )
Out[22]:
GAP: <An object in The positively graded category of The representation category of SymmetricGroup( [ 1 .. 4 ] )>
In [23]:
SW[0]
Out[23]:
$$𝟙$$
In [24]:
SW[1]
Out[24]:
$$χ$$
In [25]:
SW[2]
Out[25]:
$$ρ \oplus ν \oplus 𝟙$$
In [26]:
SW[3]
Out[26]:
$$σ \oplus 2 \cdot χ \oplus ν$$
In [27]:
SW[4]
Out[27]:
$$χ \oplus 2 \cdot ρ \oplus 2 \cdot ν \oplus 2 \cdot 𝟙$$

However, computing the multiplications $\mu^{i,j}: \mathrm{S}^i W \otimes \mathrm{S}^j W \to \mathrm{S}^{i+j} W$ goes beyond the augmented character table and needs the associator and braiding of $\mathrm{srep}_k(S_4)_+$. For computing them we currently still need an explicit irreducible representation affording each irreducible character.

In [28]:
χ¹ = InternalElement( SW, χ, 1, 1 )
Out[28]:
$$\left( \begin{array}{r} 1 \end{array} \right) χ^{\{1\}}$$
In [29]:
ρ² = InternalElement( SW, ρ, 2, 1 )
Out[29]:
$$\left( \begin{array}{r} 1 \end{array} \right) ρ^{\{2\}}$$
In [30]:
ν² = InternalElement( SW, ν, 2, 1 )
Out[30]:
$$\left( \begin{array}{r} 1 \end{array} \right) ν^{\{2\}}$$
In [31]:
ν³ = InternalElement( SW, ν, 3, 1 )
Out[31]:
$$\left( \begin{array}{r} 1 \end{array} \right) ν^{\{3\}}$$

Computing products of elements in $(\mathrm{S}W)^\in$ we get in primitive decomposition:

In [32]:
χ¹ * χ¹
Out[32]:
$$\left( \begin{array}{r} 1 \end{array} \right) ρ^{\{2\}} + \left( \begin{array}{r} 1 \end{array} \right) ν^{\{2\}} + \left( \begin{array}{r} 1 \end{array} \right) 𝟙^{\{2\}}$$
In [33]:
χ¹ * (χ¹ * χ¹)
Out[33]:
$$\left( \begin{array}{r} 1 \end{array} \right) σ^{\{3\}} + \left( \begin{array}{rr} 5 & 3 \end{array} \right) χ^{\{3\}} + \left( \begin{array}{r} 3 \end{array} \right) ν^{\{3\}}$$
In [34]:
χ¹ * (χ¹ * χ¹) == (χ¹ * χ¹) * χ¹
Out[34]:
true
In [35]:
χ¹ * ρ²
Out[35]:
$$\left( \begin{array}{rr} \cdot & 2 \end{array} \right) χ^{\{3\}} + \left( \begin{array}{r} 1 \end{array} \right) ν^{\{3\}}$$
In [36]:
χ¹ * ρ² == ρ² * χ¹
Out[36]:
true
In [37]:
χ¹ * ν²
Out[37]:
$$\left( \begin{array}{r} 1 \end{array} \right) σ^{\{3\}} + \left( \begin{array}{rr} 8 & \cdot \end{array} \right) χ^{\{3\}} + \left( \begin{array}{r} 2 \end{array} \right) ν^{\{3\}}$$
In [38]:
χ¹ * ν² == ν² * χ¹
Out[38]:
true
In [39]:
χ¹ * (χ¹ * (χ¹ * χ¹))
Out[39]:
$$\left( \begin{array}{r} -2 \end{array} \right) χ^{\{4\}} + \left( \begin{array}{rr} 3 & -31 \end{array} \right) ρ^{\{4\}} + \left( \begin{array}{rr} 5 & 6 \end{array} \right) ν^{\{4\}} + \left( \begin{array}{rr} 5 & 3 \end{array} \right) 𝟙^{\{4\}}$$
In [40]:
(χ¹ * χ¹) * (χ¹ * χ¹)
Out[40]:
$$\left( \begin{array}{r} 8 \end{array} \right) χ^{\{4\}} + \left( \begin{array}{rr} \frac{9}{4} & \frac{43}{4} \end{array} \right) ρ^{\{4\}} + \left( \begin{array}{rr} -16 & -6 \end{array} \right) ν^{\{4\}} + \left( \begin{array}{rr} -7 & \frac{9}{8} \end{array} \right) 𝟙^{\{4\}}$$
In [41]:
((χ¹ * χ¹) * χ¹) * χ¹
Out[41]:
$$\left( \begin{array}{r} 26 \end{array} \right) χ^{\{4\}} + \left( \begin{array}{rr} 3 & -31 \end{array} \right) ρ^{\{4\}} + \left( \begin{array}{rr} 5 & 6 \end{array} \right) ν^{\{4\}} + \left( \begin{array}{rr} 5 & 3 \end{array} \right) 𝟙^{\{4\}}$$
In [42]:
(χ¹ * χ¹) * ρ²
Out[42]:
$$\left( \begin{array}{r} 4 \end{array} \right) χ^{\{4\}} + \left( \begin{array}{rr} \frac{5}{4} & -\frac{9}{4} \end{array} \right) ρ^{\{4\}} + \left( \begin{array}{rr} -4 & -4 \end{array} \right) ν^{\{4\}} + \left( \begin{array}{rr} \cdot & \frac{1}{8} \end{array} \right) 𝟙^{\{4\}}$$
In [43]:
χ¹ * (χ¹ * ρ²)
Out[43]:
$$\left( \begin{array}{r} -2 \end{array} \right) χ^{\{4\}} + \left( \begin{array}{rr} 2 & -12 \end{array} \right) ρ^{\{4\}} + \left( \begin{array}{rr} -1 & -1 \end{array} \right) ν^{\{4\}} + \left( \begin{array}{rr} \cdot & 2 \end{array} \right) 𝟙^{\{4\}}$$
In [44]:
(χ¹ * χ¹) * ν²
Out[44]:
$$\left( \begin{array}{r} 4 \end{array} \right) χ^{\{4\}} + \left( \begin{array}{rr} \cdot & 16 \end{array} \right) ρ^{\{4\}} + \left( \begin{array}{rr} -8 & -1 \end{array} \right) ν^{\{4\}} + \left( \begin{array}{rr} -4 & \cdot \end{array} \right) 𝟙^{\{4\}}$$
In [45]:
χ¹ * ν³
Out[45]:
$$\left( \begin{array}{r} 4 \end{array} \right) χ^{\{4\}} + \left( \begin{array}{rr} \cdot & -12 \end{array} \right) ρ^{\{4\}} + \left( \begin{array}{rr} 1 & 1 \end{array} \right) ν^{\{4\}}$$
In [46]:
χ¹ * ν³ == ν³ * χ¹
Out[46]:
true

We see from the above computations that

In [47]:
χ¹ * (χ¹ * (χ¹ * χ¹)) ≠ ((χ¹ * χ¹) * χ¹) * χ¹
Out[47]:
true

so, as expected, the product is neither associative nor commutative as each of the two properties would imply equality.

In [ ]: