Schur functions and semistandard tableaux
Connect tableau weights, Jacobi–Trudi determinants and stable Schur bases with explicit row and column conventions.
On this page
The question for today
What you’ll understand
- Enumerate semistandard tableaux using English diagrams.
- Compute a Schur polynomial from its complete-function determinant.
- Separate monomial coefficients from tableau totals and finite-variable vanishing.
Schur functions combine a manifestly positive counting rule with determinant identities that can involve cancellation. We use English Young diagrams: the longest row is on top, rows extend rightward, and columns run downward. A semistandard tableau has positive integer entries weakly increasing along rows and strictly increasing down columns. These conventions determine the generating function. Reversing an inequality or transposing a diagram changes which symmetric function the same-looking array describes.
Weights count entries rather than positions
For shape λ, assign a variable xi to every entry i and multiply over all boxes. The Schur function sλ is the sum of these weights over all semistandard tableaux of shape λ. In N variables, entries are restricted to 1 through N. A column longer than N cannot be strictly increasing, so the specialization vanishes when λ has more than N rows. By contrast, a single row can have arbitrarily many boxes because repeated entries along a row are allowed.
Collecting equal monomials yields nonnegative integers called Kostka numbers. A coefficient counts tableaux with one specified content vector, not every rearrangement of that vector at once. Symmetry ensures that permuted contents have equal counts; proving this requires a weight-switching argument or an equivalent algebraic construction. It does not follow merely from the tableau definition looking symmetric, because the order on the entry alphabet enters the row and column conditions asymmetrically.
A determinant computes the same generating function
The Jacobi–Trudi theorem expresses sλ as the determinant with entry h at index λi−i+j, using at least as many rows as the length of λ and padding λ with zeros if extra rows are used. Negative-index complete functions are zero and h0 is one. This is an equality in the stable ring over Z. Its determinant proof can be organized by intersecting lattice paths: a sign-reversing involution cancels intersecting path families, leaving the nonintersecting ones that encode tableaux. The theorem is used here with that interpretation rather than proved in full.
The one-row determinant gives sr=hr, while the one-column version gives s at shape (1 repeated r times) equal to er. These boundary cases are valuable convention checks. More generally, Schur functions form an integral basis in every degree. Therefore a symmetric function can be Schur-positive even if a determinant used to calculate it has minus signs. Positivity concerns its final basis coordinates and the underlying combinatorial objects, not the syntax of one formula.
Content and dimension are different specializations
Setting every variable equal to one counts all admissible tableaux on that finite alphabet. In representation theory over C, this number is also the dimension of the corresponding polynomial irreducible general linear representation when the shape has at most N rows. That interpretation is an additional theorem; the tableau count is already defined without representation theory. Setting selected variables to zero instead restricts allowed entries and can reveal that a nonzero stable function has disappeared under a small-alphabet specialization.
For shape (2,1) on a two-letter alphabet, strictness forces the left column to be 1 above 2. Only the top-right box remains free, and it may be either 1 or 2. This makes the stable distinctions tangible: the two monomials have different contents even though their all-ones values coincide. Standard tableaux are a separate specialization of the combinatorics, using each label exactly once; they are not obtained simply by setting every xi to one.
Work it through
Two tableaux behind a determinant
Calculate s₂₁(x,y) by both tableaux and Jacobi–Trudi, then evaluate at x=2 and y=3.
The first column must be 1 above 2. Top rows [1,1] and [1,2], each with bottom entry 2, give weights x²y and xy²; there are no other tableaux.
In two variables h₂=x²+xy+y², h₁=x+y and h₃=x³+x²y+xy²+y³. Subtracting h₃ from h₂h₁ leaves exactly x²y+xy².
At x=2,y=3 the two weights are 12 and 18. Their sum is 30, agreeing with 19·5−65 from the determinant calculation.
s₂₁(x,y)=x²y+xy² and s₂₁(2,3)=30; at the all-ones alphabet it counts exactly two tableaux.
The living notebook
Explore the model
Interactive lab · probability
Word counts exceed constrained fillings
Pause, reset, then set p=0.5 and n=3. Predict probability 3/8 for exactly one success, representing three binary words with one 2 and two 1s. The shape (2,1) permits only one tableau with that content, of weight x₁²x₂. Compare the three unconstrained arrangements with the single manually enumerated filling. The bars enumerate words, not tableaux or Schur coefficients.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA tableau condition selects a structured subset of all arrangements with the same content.
Pause & reason
Check your understanding
Your turn
Try it yourself
Find s₂₂(x,y) and s₁₁₁(x,y), explaining their different behavior rather than only evaluating determinants.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Examine the two columns of shape (2,2), and then the height of the column of shape (1,1,1).
Compare with the solution
Every height-two column over {1,2} is forced to be 1 above 2. Hence shape (2,2) has the unique filling with top row [1,1] and bottom row [2,2], giving x²y². Shape (1,1,1) has a column of height three and cannot be strictly increasing with only two possible entries, so its Schur polynomial is zero. The stable function s₁₁₁=e₃ is still nonzero before specialization.
Keep the idea
Take it with you
- Declare English diagrams, weak rows and strict columns before counting.
- Jacobi–Trudi determinants and tableau sums compute the same integral basis.
- Finite-alphabet vanishing does not imply that the stable Schur function is zero.
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