Doubts about graded algebra.

318 Views Asked by At

In the last days I've been studying the tensor algebra $T(V)$ of a vector space $V$ over the field $K$ and I've realised that what I'm not understanding hasn't to do with tensor products, but rather with graded algebras built from direct sums.

Following the definition of wikipedia, a graded algebra is a graded vector space that is also a graded ring, that is, a vector space $V$ that can be decomposed as a direct sum

$$V=\bigoplus_{n\in\mathbb{N}}V_n$$

and with the property that if $\odot$ denotes the multiplication, then $V_n\odot V_m \subseteq V_{n+m}$.

That's fine, however what makes me in doubt is the following: suppose we have a collection of vector spaces $\{V_i, i\in\mathbb{N}\}$ and we know how to define a multiplication $\odot: V_n\times V_m\to V_{n+m}$ that satisfies the axioms of the multiplication of a ring.

Then, we can build the vector space

$$V=\bigoplus_{n\in\mathbb{N}}V_n,$$

which is a graded vector space, because if $i_n : V_n\to V$ is the canonical injection then $V$ is the internal direct sum of all $i_n(V_n)$ for $n\in \mathbb{N}$.

But how does one define multiplication in $V$? We know how to define multiplication for each two $V_n$ and $V_m$, but how can one use this to define multiplication in $V$? The elements of $V$ are sequences $(v_i)$ where each $v_i \in V_i$ and just finitely many of those $v_i$ are nonzero, but I don't know what to do with this together with the maps $\odot$ to define the multiplication of $V$ turning it into a graded ring also.

How is that usually done?

Thanks very much in advance!

2

There are 2 best solutions below

1
On

In the most "obvious" way possible: extending the multiplication via linearity.

Treat elements of $\bigoplus V_n$ as (finite) formal linear combinations of elements from the $V_n$'s, and to multiply any two of these formal sums, simply invoke distributivity (i.e. the "FOIL" rule).

0
On

You can define multiplication by $$(\vec v \odot \vec w)_i = \sum_{p+q=i} v_p \odot w_q$$ where $\vec v = (v_j : j \ge 0)$ with $v_j \in V_j$ and so on.

It's a bit like what happens when you multiply powers of a variable $x$: $$\left( \sum_{i=0}^{\infty} a_ix^i \right) \left(\sum_{j=0}^{\infty} b_jx^j \right) = \sum_{n=0}^{\infty} \sum_{i+j=n} a_ib_jx^n$$