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A naive counting of the cardinal of following basis of $\operatorname{Sym}^n(V)$ (cf. the original question for the notation and then link p.33 in this question) $$ \Big\lbrace S(e_{i_1}\otimes \cdots \otimes e_{i_n}),\ 1\leq i_1 \leq i_2 \leq \cdots \leq i_n \leq d \Big\rbrace \tag{1} \label{1}$$ yields the following formula (for the first equality cf. for example here or below) $$\operatorname{dim}\big(\operatorname{Sym}^n(V) \big) = { d+n-1 \choose n} \stackrel{??}{=}\sum_{k=1}^{\min(d,n)} {d \choose k} S(n,k)\tag{2} \label{2}$$ involving Stirling numbers of the second kind $S(n,k)$ which is just the number of partitions of a set with $n$ elements in $k$ different subsets.

((\ref{2}) is also the number of unitary monomial in $d$ variables of degree $n$ or the number of partial derivatives of order $n$ of a function of $d$ variables.)

As a rule, "if one does not see this formula anywhere, it is probably wrong" ... but here is a

Proof??: There are

  • ${d \choose 1}$ possible vectors of the form $e_{i}\otimes \cdots \otimes e_{i}$,
  • ${d \choose 2}$ possible choices of a pair $(e_i, e_j)$ of different basis vector. But a basic tensor may "contain" only 1 $e_i$ and the rest of $e_j$, or two $e_i$ and the rest of $e_j$, or three $e_i$ and the rest of $e_j$ etc. Each possibility correspond to a partition of $n$ in two (non trivial) subsets and there are $S(n,2)$ partitions.
  • ${d \choose 3}\times S(n,3)$ possible vectors "containing" exactly three different basis vectors $e_{i}, e_j$ and $e_k$.
  • etc. ... It stops at $k=\min(d,n)$.

The first equality is given by the usual trick (e.g. Problem 2. p.12 in PDEs (2010) by Lawrence Evans). Consider $d+n-1$ boxes $$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline & & & & & & & & &\\ \hline \end{array}$$ out of which one chooses $n\ $ (chosen boxes $\bullet$; rest filled with $\vert$) $$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \vert & \vert& \bullet & \vert & \bullet & \bullet & \vert & \vert & \bullet & \vert \\ \hline \end{array}$$ $$ \begin{array}{cccccccccccc} \vert & \vert& \bullet & \vert & \bullet & \bullet & \vert & \vert & \bullet & \vert \end{array} \tag{3} \label{3}$$ There are of course ${ d+n-1 \choose n}$ possibilities each of which corresponds to a choice of $(i_1, \cdots, i_n),\ 1\leq i_1 \leq i_2 \leq \cdots \leq i_n \leq d $ as in (\ref{1}): each separator, delimiter $\vert$ should be thought as the side of a new sequence of boxes: i.e. (\ref{3}) should be thought as $$ \begin{array}{cccccccccccccccc} 1 &\vert & 2 & \vert & \bullet & \vert & \bullet & \bullet & \vert & 5 & \vert & \bullet & \vert & 7 \end{array} \tag{4} \label{4}$$ In this example where $d=7$ and $n=4$, (\ref{4}) corresponds to $(i_1, \cdots, i_4)=(3,4,4,6)$

Noix07
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    The Stirling numbers are for partitioning distinct objects into non-distinct boxes. For this problem, you are partitioning non-distinct objects (which determine how much each basis vectors appears) into distinct boxes (the basis vectors, $e_i$ and $e_j$). To correct your formula, you need to replace $S(n,k)$ with $\binom{n-1}{k-1}$. – Mike Earnest Jan 21 '20 at 16:08

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The formula $$ \sum_{k=1}^{\min(d,n)}\binom{d}k\color{red}{S(n,k)}=\binom{d+n-1}{n} $$ is incorrect. The correct formula is $$ \sum_{k=1}^{\min(d,n)}\binom{d}k\color{green}{\binom{n-1}{k-1}}=\binom{d+n-1}{n} $$ Here is why the second is correct. After choosing which of the $k$ basis vectors will comprise the tensor product $e_{i_1}\otimes \dots\otimes e_{i_k}$, you need to choose the exponents of each factor $e_{i_j}$, that is, how many times each factor is repeated. Each factor must be repeated at least once, and the sum of the multiplicities must equal $n$.

There is a clever way to approach this problem; take a row of $n$ identical dots. There are $n-1$ spaces between adjacent pairs of these dots. Choose $k-1$ of these spaces, and place a vertical bar in the spaces; this can be done in $\color{green}{\binom{n-1}{k-1}}$ ways. These bars divide the $n$ dots into $k$ consecutive groups. The number of dots in each group represents the multiplicity of each basis vector.

The Stirling numbers do not apply here. $S(n,k)$ counts the number of ways to put $n$ distinct objects into $k$ identical boxes so that each box is nonempty. For this problem, you are putting identical balls into distinct boxes; each box represents one of the basis vectors of $V$, and the number of balls in each box determines the multiplicity of that vector in the tensor product. You can check directly your formula does not work when $d=2$, as $$ \sum_{k=1}^2\binom{2}kS(n,k)=2\cdot S(n,1)+1\cdot S(n,2)=2\cdot 1+1\cdot (2^{n-1}-1)\neq \binom{2+n-1}{n}=n+1 $$

Mike Earnest
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