Vtyjkyuk

We say a function is $k$-linear if it takes $k$ values as input and is linear with respect to each of them. For example, determinant is a $n$-linear function. (If the matrix is $n times n$)

A tensor is a function $T:V times Vtimes Vtimes dotstimes Vto mathbb R$ ($k$ vectors taken as input) such that $T$ is $k$-linear. (Its linear with respect to each of its $k$ inputs) ($V$ is a vector space)

A symmetric tensor is a tensor that is invariant under a permutation of its vector arguments. Meaning:

$T(v_1,v_2,dots,v_r)=T(v_sigma(1),v_sigma(2),dots,v_sigma(r))$ For each permutation $sigma$ of the symbols $1,2,dots,r$.

We call $Sym^k(V)$ the vector space of all symmetric $k$-tensors on vector space $V$.

If $T$ is a $m$-tensor and $S$ is a $n$-tensor, then $T otimes S$ is a $m+n$-tensor such that for each $(v_1,dots,v_m,v_m+1,dots,v_m+n)$ $T otimes S(v_1,dots,v_m,v_m+1,dots,v_m+n)=T(v_1,dots,v_m)S(v_m+1,dots,v_m+n)$.

Now we want to find a basis for this vector space.

I know that the basis should consist of something related to the sum of tensor products of the elements of the basis of the dual space of $V$ (called $V^*$). But i can't see how. The complete Question is written below:

Let $V$ be an $n$-dimensional vector space. Compute the dimension of $Sym^k(V)$. (It can be also found here Page 33)

I know that every $k$-tensor can be written as a linear combination of $e^i_1otimes e^i_2otimesdotsotimes e^i_k$ such that $1 le i_1,dots,i_kle n$. But i don't know which members should be eliminated to form a basis for just the symmetric $k$-tensors (Not all $k$-tensors).

Note: For example, I know that if $k=3$, A member of the basis of $Sym^3(V)$ is $e^1otimes e^2 otimes e^1+e^2otimes e^1 otimes e^1+e^1otimes e^1 otimes e^2$. But i don't know why! I want explanation. I have the answer... The dimension of $Sym^k(V)$ is $n+k-1 choose k$. My problem is that i don't know why this sum and why the coefficient should be the same for those elements in the sum.

I assume that $e_1,dots,e_n$ is a basis of $V$ and that $e^1,dots,e^n$ the associated dual basis of $V^*$.

First, let's consider the case of arbitrary (not necessarily symmetric) tensors. We note that, by linearity,
$$
T(v^(1), dots, v^(k)) =
Tleft( sum_i=1^n v^(1)_i e_i, dots, sum_i=1^n v^(k)_i e_i right) =
Tleft( sum_i_1=1^n v^(1)_i_1 e_i, dots, sum_i_k=1^n v^(k)_i_k e_i_k right) = \
sum_i_1=1^n cdots sum_i_k=1^n v^(1)_i_1 cdots v^(k)_i_k Tleft(e_i_1, dots, e_i_k right)
$$
Now, define the tensor $tilde T$ by
$$
tilde T = sum_i_1=1^n cdots sum_i_k=1^n Tleft(e_i_1, dots, e_i_k right) e^i_1 otimes cdots otimes e^i_k
$$
Prove that $tilde T(v^(1),dots,v^(k)) = T(v^(1),dots,v^(k))$ for any $v^(1),dots,v^(k)$. That is, $tilde T = T$. We've thus shown that any (not necessarily symmetric) $k$-tensor can be written as a linear combination of $e^i_1 otimes cdots otimes e^i_k$.

The same applies for symmetric tensors. However, if $T$ is symmetric, then
$$
Tleft(e_i_1, dots, e_i_k right) =
Tleft(e_sigma(i_1), dots, e_sigma(i_k) right)
$$
for any permutation $sigma$. Thus, we may regroup the above sum as
$$
T = tilde T = sum_i_1=1^n cdots sum_i_k=1^n Tleft(e_i_1, dots, e_i_k right) e^i_1 otimes cdots otimes e^i_k =
\
sum_1 leq i_1 leq cdots leq i_k leq n ;
frac 1alpha(i_1,dots,i_k)sum_sigma in S_k Tleft(e_sigma(i_1), dots, e_sigma(i_k) right)
e^sigma(i_1) otimes cdots otimes e^sigma(i_k) =
\
sum_1 leq i_1 leq cdots leq i_k leq n ;
frac 1alpha(i_1,dots,i_k)sum_sigma in S_k Tleft(e_i_1, dots, e_i_k right)
e^sigma(i_1) otimes cdots otimes e^sigma(i_k) =
\
sum_1 leq i_1 leq cdots leq i_k leq n
frac 1alpha(i_1,dots,i_k)
Tleft(e_i_1, dots, e_i_k right)
underbracesum_sigma in S_k e^sigma(i_1) otimes cdots otimes e^sigma(i_k)_textbasis element for Sym^k(V)
$$
Thus, we have expressed $T$ as a linear combination of the desired basis elements.

$alpha(i_1,dots,i_k)$ counts the number of times any element $(sigma(i_1),dots,sigma(i_n))$ appears in the summation over $sigma in S_n$. As the comment below points out, we have
$$
alpha(i_1,dots,i_k) = m_1! cdots m_n!
$$
where $m_j$ is the multiplicity of $j in 1,dots,n$ in the tuple $(i_1,dots,i_k)$.

Let's consider $mathbbR^2$ with standard basis $e_1,e_2$.

If $T$ is a symmetric tensor $T:mathbbR^2timesmathbbR^2timesmathbbR^2tomathbbR$, then we can group basis vectors $e^i_1otimes e^i_2otimes e^i_3$ of the tensor power $(mathbbR^2)^otimes 3$ according to whether or not $T$ must send them to the same value:

$$ beginarrayrrrr e_1otimes e_1otimes e_1 \ hline e_1otimes e_1otimes e_2 & e_1otimes e_2otimes e_1 & e_2otimes e_1otimes e_1 \ hline e_1otimes e_2otimes e_2 & e_2otimes e_1otimes e_2 & e_2otimes e_2otimes e_1 \ hline e_2otimes e_2otimes e_2 endarray $$

In other words, the values $T$ takes on any tensor can be determined as long as we know what values $T$ takes on

1. $e_1otimes e_1otimes e_1$,


2. $e_1otimes e_1otimes e_2$,


3. $e_1otimes e_2otimes e_2$,


4. $e_2otimes e_2otimes e_2$.

These are precisely the basis elements $e_i_1otimes e_i_2otimes e_i_3$ with $i_1le i_2le i_3$.

Conversely, given any four values $a,b,c,d$ we can arrange for $T$ to take these values on the above basis vectors by writing out

$$ beginarraylll T & = & a(e^1otimes e^1otimes e^1) \ &+ & b(e^1otimes e^1otimes e^2+e^1otimes e^2otimes e^1+e^2otimes e^1otimes e^1) \ & + & c(e^1otimes e^2otimes e^2+e^2otimes e^1otimes e^2+e^2otimes e^2otimes e^1) \ & + & d(e^2otimes e^2otimes e^2), endarray $$

Generalize.