If I understand that article correctly, the symmetric algebra $S(\mathbb R^n)$ is (isomorphic to) the algebra of polynomials with $n$ variables. As a vector space, it is infinite-dimensional, despite finite $n$ (consider $\{1,x,x^2,x^3,x^4,\cdots\}$ as basis vectors). It has a multiplication that is commutative and associative.

Can this be described in terms of the exterior algebra $\Lambda(\mathbb R^\infty)$?

(By $\mathbb R^\infty$ I mean $\mathbb R^\mathbb N$, the space of Real-valued sequences... or perhaps $\ell^\infty$; I'm not familiar enough with these to know what "ought" to be used.)

For $n=2$, consider these infinite sums of bivectors, where $e_ie_j=e_i\wedge e_j$ is the exterior product, and all $e_i$ and $f_i$ are linearly independent:

$$x = e_1e_2+e_3e_4+e_5e_6+\cdots$$ $$y = f_1f_2+f_3f_4+f_5f_6+\cdots$$

The exterior product is associative, and anticommutative for vectors, but commutative for bivectors: $(e_1e_2)(e_3e_4) = -e_1e_3e_2e_4 = e_1e_3e_4e_2 = -e_3e_1e_4e_2 = (e_3e_4)(e_1e_2)$. So we can write the powers of $x$ as $$x^2 = (e_1e_2)(e_1e_2)+(e_1e_2)(e_3e_4)+(e_3e_4)(e_1e_2)+(e_3e_4)(e_3e_4)+\cdots$$ $$= 0+2(e_1e_2e_3e_4)+2(e_1e_2e_5e_6)+2(e_3e_4e_5e_6)+\cdots$$ $$x^3 = 6(e_1e_2e_3e_4e_5e_6)+6(e_1e_2e_3e_4e_7e_8)+\cdots$$ $$x^4 = 24(e_1e_2e_3e_4e_5e_6e_7e_8)+\cdots$$

and the product $xy$ as

$$xy = (e_1e_2f_1f_2)+(e_1e_2f_3f_4)+(e_3e_4f_1f_2)+(e_3e_4f_3f_4)+\cdots$$ $$= yx$$

and so on.

It should be clear that all of these terms $\{1,x,y,x^2,xy,y^2,x^3,x^2y,\cdots\}\subset\Lambda(\mathbb R^\infty)$ are linearly independent, as they are in the ordinary polynomial space. So, is the algebra generated by these bivectors isomorphic to the 2-variable polynomial algebra $S(\mathbb R^2)$?

Are there problems with infinity, or anything else I'm missing?

  • 1
    $\begingroup$ Your infinities are shaky. Neither $x$ nor $y$ lies in $\Lambda\left(\mathbb{R}^\infty\right)$ for any reasonable meaning of $\mathbb{R}^\infty$; only finite linear combinations are allowed in an exterior algebra. However, the question is interesting and deserves the tweaks that need to be made for it to make sense. For example, you can define $\mathbb{R}^\infty$ to be the graded vector space with basis $x_1, x_2, x_3, \ldots$, with each $x_i$ being homogeneous of degree $i$. Then, the exterior algebra $\Lambda\left(\mathbb{R}^\infty\right)$ is a connected graded algebra ... $\endgroup$ – darij grinberg Aug 9 '18 at 17:34
  • 1
    $\begingroup$ ... ("connected" here just means that the $0$-th graded component is spanned by $1$), and thus has a completion with respect to the grading. This completion does contain $x$ and $y$, and more generally contains wedges of the forms $\sum_{k=0}^{\infty} x_{kn+1} \wedge x_{kn+2} \wedge \cdots \wedge x_{kn+n}$ for all $n > 0$. When $n$ is even, these wedges are even and thus commute with each other, so it makes sense to ask whether they form a polynomial ring. $\endgroup$ – darij grinberg Aug 9 '18 at 17:36
  • 1
    $\begingroup$ Let $V = \bigoplus_{n=0}^{\infty} V_n$ be an $\mathbb{N}$-graded $R$-module (with $R$ being a commutative ring). Then, the completion of $V$ with respect to the grading is defined to be the topological $R$-module $\widehat{V} := \prod_{n=0}^{\infty} V_n$, whose topology is defined in such a way that a net $\left(v_i\right)_{i\in I}$ converges to an element $v$ if and only if for each nonnegative integer $n$, every sufficiently large $i \in I$ satisfies $\pi_n\left(v_i\right) = \pi_n\left(v\right)$. Here, $\pi_n$ denotes the canonical projection $\widehat{V} \to V$. (If you ... $\endgroup$ – darij grinberg Aug 9 '18 at 23:25
  • 1
    $\begingroup$ ... don't know what a net is, read "sequence" for "net", but beware that convergence of sequences alone doesn't quite define a topology. It's mostly a technicality that doesn't bite you unless you do something weird.) Informally speaking, $\widehat{V}$ is what becomes of $V$ if we introduce infinite sums, as long as the addends of these sums have to "recede into higher and higher degrees". Now, if $V$ is an $R$-algebra, then it is easy to define an $R$-algebra structure on $\widehat{V}$ as well. So you can complete any graded $R$-algebra with respect to its grading. I think that ... $\endgroup$ – darij grinberg Aug 9 '18 at 23:27
  • 1
    $\begingroup$ (1): Yes, sorry. (2): That's one way to define it. (I find it somewhat more natural to replace "$\pi_n\left(v_i-v_j\right)=0$" by "$\forall m \leq n, \pi_m\left(v_i-v_j\right) = 0$"; but this is equivalent.) (3): It is the sum of the grades of all vectors in a product. (4): Yes, something like that is my intuition too. $\endgroup$ – darij grinberg Aug 10 '18 at 1:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.