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The Schur polynomials satisfy many, many identities and there is a whole book about them. I think the easiest way is with the Vandermonde Determinant. $$s_{3,1,1}(a,b,c) = \frac{\left|\begin{array}{ccc} a^5 & b^5 & c^5 \\\\ a^2 & b^2 & c^2 \\\\ a & b & c\end{array} \right|}{(a - b)(b - c)(c - a)} = \left|\begin{array}{ccc} e_3 & e_2 & e_1 \\\\ e_2 & e_1 & e_0 \\\\ e_3 & e_2 & e_1\end{array} \right| = \left|\begin{array}{ccc} h_3 & h_2 & h_1 \\\\ h_2 & h_1 & h_0 \\\\ h_3 & h_2 & h_1\end{array} \right|$$ Where the $e_k$ are elementary symmetric polynomials and the $h_k$ are the homogeneous symmetric polynomials (the first is linear in each variable, the other isn't.) (3,1,1) is its own transpose.

Are there similar, easy to compute ways to find skew-Schur polynomials? For example, how to find $s_{(3,2,1)\backslash(1)}(x_1, x_2, x_3)$ - which is three squares along a diagonal the W penomino?

I know it is also possible to get (skew-)Schur polynomials as the matrix elements of a certain propagator acting on free fermions and using the Wick formula. For example, here.

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It looks to me like (3,1,1)/(2,1) is two squares on the diagonal with a gap in the middle. Perhaps a combinatorialist can correct me if I'm wrong, but I think that if you can split your skew diagram as the union of two with no common side of a box, then you just get the product of the Schur functions for the two pieces. So $s_{(3,1,1)/(2,1)}$ would be $e_1^2$. – Ben Webster Apr 25 '10 at 16:24
Also, how do you know that the Schur polynomials appear in that propogator if you don't have a source? It's not really very productive to throw out a random comment like that with no context on you've done to investigate on your own. Were you told about it in person? Did you read something that mentioned them in passing? – Ben Webster Apr 25 '10 at 16:27
You might also look at Stanley v.2: – Steve Huntsman Apr 25 '10 at 17:13
@ Ben. Is there a pun on the word "source", since we're talking about propagators? I've added a literature reference. – john mangual Apr 25 '10 at 18:13
Describing Macdonald's book as «a book about identities involving Schur polynomials» is leaving 90% of the book in the background! – Mariano Suárez-Alvarez Apr 26 '10 at 2:45

From Macdonald's book you have the identities that Igor Pak mentions: $$s_{\lambda/\mu}=\det(e_{\lambda_i'-\mu_j'-i+j})_{1\le i,j \le n}=\det(h _{\lambda_i-\mu_j-i+j}) _{1\le i,j\le n}$$ The skew Schur functions are defined by the relation $\langle s _{\lambda/\mu},s _\nu\rangle=\langle s _{\lambda},s _{\mu}s _{\nu}\rangle$ (that's the Hall inner product) and therefore they satisfy $$s _{\lambda/\mu}=\sum _{\nu}c^{\lambda} _{\mu \nu} s _{\nu}$$ where $c^{\lambda} _{\mu \nu}$ are the Littlewood-Richardson coefficients, which can be computed either using the Littlewood-Richardson rule or other methods like Zelevinsky's pictures or Remmel and Whitney's reverse numbering (explained in Fulton's book "Young tableaux").

Now, I don't see how the fermionic definitions help in computing (skew) Schur functions, in a more practical way. The fermion Fock space is naturally isomorphic to the space of skew-symmetric polynomials. It's worth mentioning the paper "The Littlewood-Richardson rule and the boson-fermion correspondence" by Baker where it is pointed out how this method could be used to compute skew Schur functions, and the paper "A combinatorial generalization of the boson-fermion correspondence" by Lam which explores what happens if one decides to replace the Fermionic Fock space with another representation of the Heisenberg algebra. However I don't think that the boson/fermion language implies new formulas, or that this physical intuition is useful at the level of simply computing these functions. You'll be doing what is already in Macdonald, the boson-fermion correspondence is just essentially multiplying by a Vandermonde determinant etc.

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As this is a representation theory question, the connection to affine Hecke algebras deserves a few more words. The (degenerate) affine Hecke algebra, $H_d$ is isomorphic as a vector space to $\mathbb{C}[x_1,\ldots,x_d]\otimes\mathbb{C}S_d$, so that $\mathbb{C}[x_1,\ldots,x_d]\otimes 1$ and $1\otimes\mathbb{C}S_d$ are subalgebras (which we identify with $\mathbb{C}[x_1,\ldots,x_d]$ and $\mathbb{C}S_d$ respectively) and satisfy the mixed relations are $x_js_i=s_ix_j$ if $j\neq i,i+1$, and $x_{i+1}s_i=s_ix_i+1$. There is an equivalent story in the nondegenerate case as well.

Now, the subalgebra $\mathbb{C}[x_1,\ldots,x_d]$ is maximal commutative, and given a finite dimensional $H_d$-module $M$, the Chinese remainder theorem yields a decomposition $$M=\bigoplus_{\underline{a}\in\mathbb{C}^n}M_{\underline{a}}^{\mathrm{gen}},$$ where $M_{\underline{a}}^{\mathrm{gen}}$ is the generalized $\underline{a}$-eigenspace for the action of $x_1,\ldots,x_d$ on $M$. That is, for $\underline{a}=(a_1,\ldots,a_d)$, there exists sufficiently large $N$ such that $(x_i-a_i)^N$ acts as 0 on $M_{\underline{a}}^{\mathrm{gen}}$ for all $i$. Let $M_{\underline{a}}\subset M_{\underline{a}}^{\mathrm{gen}}$ be the honest simultaneous eigenspace for the action of $x_1,\ldots,x_d$ (i.e. $x_i-a_i$ acts as 0).

First of all, it is straightforward to show that one can reduce the study of finite dimensional $H_d$-modules to so-called "integral" finite dimensional $H_d$-modules. These are the modules for which $M_{\underline{a}}^{\mathrm{gen}}\neq0$ implies $\underline{a}\in\mathbb{Z}^d$ and it is best to think of them as being "dense" among all the finite dimensional representations. Now, among the integral finite dimensional $H_d$-modules are the "calibrated" representations which satisfy $M_{\underline{i}}^{\mathrm{gen}}=M_{\underline{i}}$ for all $\underline{i}\in\mathbb{Z}^d$. These are parameterized precisely by skew tableaux and, on restriction to $\mathbb{C}S_d$, their characters are the skew Schur functions.

It is relatively easy to give a construction of these representations. Indeed, given a skew shape $\lambda/\mu$ consisting of $d$ boxes (and a "charge" $c$ which I will ignore for brevity), let $L(\lambda/\mu)$ be the $\mathbb{C}$ vector space with basis $$\lbrace v_T|T\mbox{ a standard filling of }\lambda/\mu\rbrace.$$ Define an action of $\mathbb{C}[x_1,\ldots,x_d]$ on $L(\lambda/\mu)$ by $x_i v_T=c(T_i)v_T$, where $c(T_i)$ denotes the content of the box in $T$ labeled $i$. In other words, $\mathbb{C}v_T=L(\lambda/\mu)_{(c(T_1),\ldots,c(T_d))}$.

The action of the symmetric group takes a bit more work. There is a natural action of $S_d$ on $\mathbb{Z}^d$ by place permutation, and similarly on the set of standard $\lambda/\mu$ tableaux. On the other hand, the mixed relation from the first paragraph implies that $s_jM_{\underline{i}}\subset M_{s_j\underline{i}}+M_{\underline{i}}$ for any integral finite dimensional $H_d$-module $M$, so $S_d$ cannot act on $L(\lambda/\mu)$ by permuting the entries in the tableaux $T$. To find the real action, we consider a system of intertwining operators $\phi_i=x_{i+1}s_i-s_ix_{i+1}=s_i(x_i-x_{i+1})+1$, $i=1,\ldots,d-1$. One can readily calculate that these operators satisfy the type $A$ braid relations and that $\phi_ix_j=x_{s_i(j)}\phi_i$. In particular, $\phi_jM_{\underline{i}}\subseteq M_{s_j\underline{i}}$. Note, however that $\phi_i^2=(x_i-x_{i+1})^2-1$ which may act as 0.

Since $L(\lambda/\mu)$ is calibrated, we must have $\phi_iv_T=Y_{i,T}v_{s_iT}$ for some scalars $Y_{i,T}$. It is clear that we must have $Y_{i,T}=0$ if $s_iT$ is not standard. Using the formula for $\phi_i$ gives that $$(c(T_{i+1})-c(T_{i})s_iv_T=v_T-Y_{i,T}v_{s_iT}.$$ If the left-hand-side is 0, then $T_i$ and $T_{i+1}$ lie on the same diagonals, which cannot happen in a standard tableau. Therefore, in all cases we can solve for $s_iv_T$. Comparing the calculation $$s_iv_{s_iT}=(c(T_i)-c(T_{i+1})^{-1}(v_{s_iT}-Y_{i,s_iT}v_T)$$ to that for $s_iv_T$ suggests that $Y_{i,T}$ and $Y_{i,s_iT}$ should be related. In fact, the right guess is that they are the same. A quick calculation using this guess and the quadratic relation $s_i^2=1$ gives that $$Y_{i,T}=\pm\sqrt{(c(i+1)-c(i))^{-2}-1}.$$ It remains to check the braid relations, which is a bit tedious. This will settle the sign issue ($\pm=+$). There is a nice paper of Arun Ram, "Skew shape representations are irreducible" (arxiv:math.RT/0401326) which does this.

In a recent work of Kleshchev and Ram (arxiv:0809.0557), the authors describe a new model for the combinatorics above in the wider context of Quiver Hecke Algebras (aka Khovanov-Lauda-Rouquier algebras) which are connected to categorified quantum groups. Toward the end of the article is a nice (albeit brief) discussion of the connection to skew Schur functions.

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I don't know what is a "propagator", but when it comes to symmetric polynomials, there is an easy way to learn: read Macdonald's "Symmetric functions and Hall polynomials". In this case, Section 1.5 talks about skew Schur functions, and eq. 5.4. and 5.5 give the det formula, generalizing the det formula you posted. There is a deep reason for this extension, involving degenerate affine Hecke algebras, but that's a long story. On a combinatorial level, see Gessel & Viennot paper for a bijective proof.

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