Exterior algebra error

asked 2017-11-14 04:59:31 +0200

Nat Mayer gravatar image


I'm new to Sage, and I've been having a lot of trouble constructing a particular algebra. I want to construct the exterior algebra (over $\mathbb{Q}$) on generators $w_{i,j} = w_{j,i}$ where $1 \le i < j \le n$ for some $n$ (for concreteness, say $n = 6$). I want this particular generating set so that I can define an $S_n$ action, but that's the next challenge. I've been attempting the following rough outline:

  1. Construct a vector space $V \cong \mathbb{Q}^n$, with basis ${v_i}$.
  2. Take a tensor product $V \otimes V$, with basis $w_{i,j} = v_i \otimes v_j$.
  3. Take a quotient to impose relations $w_{i,i} = 0$ and $w_{i,j} = w_{j,i}$.
  4. Take the exterior algebra on the quotient.

Several possible data structures for $V$ (FiniteRankFreeModule, VectorSpace, FreeModule) seem to fail at step 2. Are tensor products implemented for these? The most promising structure, CombinatorialFreeModule, fails at step 4 for an unknown reason. I get an error "base must be a ring or a subcategory of Rings()", even though the base is $\mathbb{Q}$.

Here's the specific code I've tried.

indices = range(1,7)
V = CombinatorialFreeModule(QQ, indices)
V2 = tensor((V,V))
w = V2.basis()
relations = []
for i in indices:
    for j in range(i+1,7)
        relations.append(w[i,j] - w[j,i])
R = V2.submodule(relations)
V3 = V2.quotient_module(R)
A = ExteriorAlgebra(V3)

The last line gives an error, "base must be a ring or a subcategory of Rings()". The command V2.base() in Rings() returns true, but I can't get around the error.

Any help would be appreciated, either in fixing this error or approaching the construction in a different way.

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Look at the doc and code of exterior algebra

sage: V3 in Rings()
sage: ExteriorAlgebra??
FrédéricC gravatar imageFrédéricC ( 2017-11-14 13:25:06 +0200 )edit

Thanks for the response FredericC. The documentation says that ExteriorAlgebra can take as input R a free module over a base ring. Can I get sage to treat $V3$ as a free module (vector space)? It knows that $V3$ lies in the category of vector spaces with basis over $\mathbb{Q}$.

If not, can you think of any other way to create an exterior algebra with bi-indexed basis, such that $w_{i,j} = w_{j,i}$?

Nat Mayer gravatar imageNat Mayer ( 2017-11-14 19:50:18 +0200 )edit

Indeed, I was wrong. Try this maybe:

sage: n = 4
sage: E=ExteriorAlgebra(QQ,['w{}{}'.format(i,j) for j in range(n) for i in range(j)])
sage: E.gens()
(w01, w02, w12, w03, w13, w23)
FrédéricC gravatar imageFrédéricC ( 2017-11-15 09:35:46 +0200 )edit