Unexpected outputs from TensorSymmetry

Questions
,
1

I have encountered what looks like a bug relating to tensor symmetry computations (or maybe I am missing something?)

Take the following example:

Table[E^(-(((i - j)/2 - p + q)^2/2)), {i, 5}, {j, 5}, {p, 3}, {q, 3}]

This array is symmetric if you swap slots 1 with 2 and 3 with 4. TensorSymmetry should give {{Cycles[{{1, 2}, {3, 4}}], 1}}. However, it doesn’t detect any symmetries at all. Checking

\[Element] Arrays[{5, 5, 3, 3}, Reals, {{Cycles[{{1, 2}, {3, 4}}], 1}}]

Also returns false. Passing to SymmetrizedArray also returns an array with no symmetries. Note that if you explicitly specify the symmetry it will work.

SymmetrizedArray[{i_, j_, p_, q_} :> E^(-(((i - j)/2 - p + q)^2/2)), {5, 5, 3, 3}, {{{2, 1, 4, 3}, 1}}]

The array you get will be symmetric and works with \[Element] and TensorSymmetry. In fact, it will pass an equality check with the non-symmetrized version, proving the array is indeed symmetric.

SymmetrizedArray[{i_, j_, p_, q_} :> E^(-(((i - j)/2 - p + q)^2/2)), 
 {5, 5, 3, 3}, {{{2, 1, 4, 3}, 1}}] == 
 Table[E^(-(((i - j)/2 - p + q)^2/2)), {i, 5}, {j, 5}, {p, 3}, {q, 3}]

Gives True, and

Symmetrize[
  Table[E^(-(((i - j)/2 - p + q)^2/2)), {i, 5}, {j, 5}, {p, 3}, {q, 3}], {{{2, 1, 4, 3}, 1}}] == 
 Table[E^(-(((i - j)/2 - p + q)^2/2)), {i, 5}, {j, 5}, {p, 3}, {q, 3}]

also gives True.

I suspect there is a bug in the symmetry finding or checking code. Here I am using exact numbers, so there shouldn’t be any tolerance issues. But if you use approximate numbers you get the same issue. This is on Wolfram Desktop 13.2.1.0 on MacOS ARM.

2

Hi Nicholas,

the tensor is definitely not symmetric

In[151]:= ((i - j)/2 - p + q)^2/2 - ((j - i)/2 - p + q)^2/2 // Simplify

Out[151]= (i - j) (-1 + q)

Regards Roland

3

It’s true that it’s not symmetric exchanging only i with j, but it is symmetric exchanging both i with j and p with q. This is what the symmetry {{2,1,4,3},1} indicates–equivalently {Cycles[{{1,2},{3,4}],1}. The symmetry you are checking for is {Cycles[{{1,2}}],1} or {{2,1,3,4},1}.

((i - j)/2 - p + q)^2/2 == ((j - i)/2 - q + p)^2/2 // Simplify

gives True

4

T is transpose-symmetric {1->2, 3->4}

In:=(T = 
   Table[ ((i - j) - p + q)^2 , {i, 4}, {j, 5}, {p, 3}, {q, 2}]) ==  
        Table[ ((j - i) - q + p)^2 , {i, 4}, {j, 5}, {p, 3}, {q, 2}]

Out:= True

Tensorsymmetry[] seems to detect a symmetry of the simultaneous transpositions in groups only if the Rank is uniform.

In:= TensorSymmetry[T]

Out:= {}

In:= T = Table[ ((i - j) - p + q)^2 , {i, 4}, {j, 4}, {p, 4}, {q, 4}] ;

In In[294]:= TensorSymmetry[T]

Out= {{Cycles[{{2, 3}}], 1}, {Cycles[{{1, 2}, {3, 4}}],  1}, {Cycles[{{1, 2, 4, 3}}], 1}}

Yoe example reduced to the index function tensor

T = Table[ ((i - j) - p + q)^2 , {i, 4}, {j, 4}, {p, 3}, {q, 3}] 

In:= TensorSymmetry[T]

Out[296]= {}

Regards Roland