What do we mean when we say the Schur functions form a basis.

Question

This has always bugged me. When we are examining symmetric functions (or polynomials if you prefer finitely many variables), we have an easy choice of basis with the monomial symmetric functions. As this basis tends to be hard to work with, we then look at the complete homogeneous symmetric functions and the elementary symmetric functions. Finally, we see that to prove things (and to link to Representation theory), it is usually best to work in the Schur basis. What I don't get is why the Schur basis is indeed a basis. It is not clear to be because we have the following relationships:

$$s_{(i)} = h_{i}$$ $$s_{\left(1^\ell\right)} = e_{\ell}$$

This clearly shows that the complete homogeneous symmetric functions and the elementary symmetric functions are special cases of the Schur functions. Would this not lead to independence issues for the Schur functions, as there are clearly more Schur functions then the ones described above, which would mean the set of Schur functions is not linearly independent?

Thanks for any help, this has driven me crazy for a while!

Answer 1

The $h_i$ are a special case of the Schur functions, but the $h_i$ by themselves do not form a basis. You would need the $h_\lambda$ for all partitions $\lambda$ to get a basis from the homogeneous symmetric functions (defined by $h_\lambda = \prod h_{\lambda_i}$), just like you need the $s_\lambda$ for all $\lambda$ to get a basis from the Schur functions. Similarly for the elementary symmetric functions.

Answer 2

I think there is some confusion here

$$s_{(i)}=h_i$$ $$s_{(1^l)}=e_l$$

This clearly shows that the complete homogeneous symmetric functions and the elementary symmetric functions are special cases of the Schur functions.

It looks like you are confusing the homogeneous symmetric function $h_k$ defined for fixed $1\le k$ as

$$h_k(x_1,\ldots , x_N)=\sum_{1\le i_1\le i_2\le\cdots\le i_k\le N} x_{i_1}x_{i_2}\cdots x_{i_k}$$

with the complete homogeneous symmetric function indexed by $\alpha$ which are given by

$$h_{\alpha}(x_1, \ldots,x_N)=\prod_{i=1}^{s}h_{\alpha_i}(x_1,\ldots ,x_N)$$

where $\alpha =(\alpha_1,\ldots ,\alpha_s)$ is a sequence of positive integers.

Similarly it appears as if have confused the elementary symmetric function $e_k$ defined for fixed $1\le k$, written

$$e_k(x_1,\ldots , x_N)=\sum_{1\leq i_1\lt i_2\lt\cdots\lt i_k \leq N}x_{i_1}x_{i_2}\cdots x_{i_k}$$

with the complete elementary symmetric function indexed by $\alpha$ which are given by

$$e_{\alpha}(x_1, \ldots,x_N)=\prod_{i=1}^{s}e_{\alpha_i}(x_1,\ldots ,x_N)$$

even though in the previous description you seem to be aware that they are different.

It is true that (ranging over all partitions $\alpha\vdash n$) the complete homogeneous symmetric functions indexed by $\alpha$ ($h_{\alpha}$) and the complete elementary symmetric functions indexed by $\alpha$ ($e_{\alpha}$) form bases (for the ring of symmetric functions, in infinitely many variables); it is not true that either of the homogeneous symmetric functions $h_k$ or the elementary symmetric functions $e_k$ form bases.

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The proof that Schur polynomials form a basis involves showing that they can be transformed from the monomial symmetric function $m_{\lambda}$ for $\lambda\vdash n$ (which form a basis) with a lower triangular Kostka matrix, thus making it invertible hence the $m_{\lambda}$ can be expressed in terms of a linear sum of Schur polynomials thus Schur polynomials also form a basis.