Exterior algebra error
Hi,
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:
- Construct a vector space $V \cong \mathbb{Q}^n$, with basis ${v_i}$.
- Take a tensor product $V \otimes V$, with basis $w_{i,j} = v_i \otimes v_j$.
- Take a quotient to impose relations $w_{i,i} = 0$ and $w_{i,j} = w_{j,i}$.
- 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:
relations.append(w[i,i])
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.
Look at the doc and code of exterior algebra
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 $w_{i,j} = w_{j,i}$?
Indeed, I was wrong. Try this maybe: