2

Here is the question I am trying to solve:

Let $A = \bigoplus_{i \geq 1}A_i$ be a graded algebra such that the vector spaces $A_i$ are all finite-dimensional. Define the Poincare series of $A$ as the formal series $$P(A) = \sum_{i \geq 0} \operatorname{dim}(A_i)t^i.$$

Prove that $$P(k\{x_1, \dots , x_n \}) = \frac{1}{1 - nt} \text{ and } P(k [x_1, \dots , x_n ]) = \frac{1}{(1 - t)^n}$$

Definition 1.

Let $X$ be a set. Consider the vector space $k\{X\}$ with basis the set of all words $x_{i_{1}} \dots x_{i_{p}}$ in the alphabet $X,$ including the empty word $\emptyset.$ A word will be called a monomial. Define the degree of the monomial $x_{i_{1}} \dots x_{i_{p}}$ as its length $p.$ Concatenation of words defines a multiplication on $k\{X\}$ by $$(x_{i_{1}} \dots x_{i_{p}})(x_{i_{p + 1}} \dots x_{i_{n}}). \quad (2.1)$$

Formula $(2.1)$ equips $k\{X\}$ with an algebra structure, called the free algebra on the set $X.$ If $X = \{x_1 , \dots , x_n \}$ we denote $k\{X\}$ by $k \{x_1, \dots , x_n\}.$

Definition 2.

$k\{x_1, \dots , x_n \}/I \cong k[x_1, \dots , x_n]$ where $I = x_i x_j - x_j x_i$ where the later is a commutative algebra.

My Question is:

1- How can I prove the formula for this $P(k\{x_1, \dots , x_n \}) = \frac{1}{1 - nt}$ (the noncommutative case)

Could someone help me in answering those questions please?

EDIT 1:

Here is an answer for my second question :

As for the free algebra $K\langle x_1,\ldots,x_n\rangle$, as Qiaochu noted, the number of words of length $k$ is $n^k$. As these words are linearly independent, the dimension of the $k$th degree component is $n^k$. So, $$P(K\langle x_1,\ldots,x_n\rangle,t)=1+nt+n^2t^2+n^3t^3+\cdots=\frac{1}{1-nt}.$$

But I am not quite sure why the first equality in the last statement correct and why we have that $|nt| < 1 $. Could someone clarify this to me please?

EDIT 2:

Also, here is my second question about this Poincare Series of a graded algebra (revisited) in case you also want to answer it.

Intuition
  • 3,043
  • You quote Qiaochu but where from? did he comment here and then delete his comment? – Anne Bauval Jan 26 '23 at 18:08
  • Since (as mentionned in your edit 2) you re-posted elswhere your 1st question (about the commutative case, and with more details), I think you should delete it from here, and concentrate here on the non-commutative case. – Anne Bauval Jan 26 '23 at 19:03
  • @AnneBauval let me include the link of the solutions – Intuition Jan 26 '23 at 19:53
  • @AnneBauval ok I will edit my question, thanks for clatifying! – Intuition Jan 26 '23 at 19:53
  • I think my problem is writing $P(A)$ for $A = k[x_1 , \dots , x_n]$ and $A = k {x_1 , \dots , x_n}$ can you help me in this please @AnneBauval – Intuition Jan 26 '23 at 21:37

1 Answers1

1

Since your "question 1" (the commutative case $k[x_1,\dots,x_n]$) is treated in your new post, let us concentrate on your "question 2": the non-commutative case $A=k\{x_1,\dots,x_n\}=k\langle x_1,\ldots,x_n\rangle.$

As mentionned in your edit 1, a basis of $A_i$ is indeed the set of words $x_{j_1}\dots x_{j_i}$ of length $i$ on the alphabet $\{x_1,\dots,x_n\},$ and the number of such words is $n^i.$ So, by definition of $P_A:$ $$P(A)=\sum_{i\ge0}n^it^i=\sum_{i\ge0}(nt)^i=\frac1{1-nt},$$ where the last equality stems from the classical identity on the formal geometric power series: $(1-X)^{-1}=\sum_{n\in\Bbb N}X^n.$

Anne Bauval
  • 49,005