Symmetric functions and standard bases
Work in the graded ring of stable symmetric functions and compare monomial, elementary, complete and power-sum coordinates.
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The question for today
What you’ll understand
- Define bounded-degree symmetric functions and finite-variable specialization.
- Convert a degree-two expression between standard bases.
- Explain which bases are integral and which require rational coefficients.
Symmetry makes a polynomial insensitive to the names of its variables, but a fixed number of variables can hide identities that disappear when another variable is added. Symmetric functions solve this bookkeeping problem by retaining every finite alphabet at once. We work first over the integers, with homogeneous degree recording total exponent. Passing to rational coefficients later permits division by integers; that change matters particularly for the power-sum basis and the character formulas built from it.
Stable degree, rather than an arbitrary infinite series
Define the ring Λ over Z as symmetric formal series in countably many variables with a finite bound on total degree. Each monomial has finite support; coefficients are invariant under every finite permutation of variables. Infinite sums of monomials within one degree are allowed, but an element of this ring has only finitely many homogeneous degrees. Thus the sum of all variables belongs to Λ, whereas the sum of one complete homogeneous function in every degree belongs only to a degree completion.
Setting all variables beyond the first N equal to zero gives a symmetric polynomial in N variables. In degree d, this specialization is injective once N is at least d: every partition of d has at most d nonzero parts and its distinct monomial orbit survives. This is a degreewise statement, not injectivity on the entire ring for one fixed N. For example, the elementary function of degree N plus one vanishes on N variables while remaining a nonzero symmetric function.
Four bases organize different counting questions
A partition λ is a weakly decreasing finite list of positive integers; its size is their sum. The monomial function mλ sums every distinct monomial whose nonzero exponents rearrange to λ, once each. Elementary er sums squarefree degree-r monomials. Complete homogeneous hr sums all degree-r monomials. Their products eλ and hλ, indexed by partitions, are integral bases, as are the monomial functions. We set e0=h0=1 and use zero for negative indices when determinant identities require them.
The power sum pr is the sum of the rth powers of the variables, and pλ is the product of its parts. These products are a basis over Q, but generally not over Z. Already e2=(p1 squared minus p2)/2 requires a denominator. No contradiction arises: the numerator has even monomial coefficients, yet its coordinates in the power-sum basis involve halves. A basis is always a basis over a declared coefficient ring; integrality cannot be inferred merely from a familiar name.
Change of coordinates preserves the object, not its interpretation
Compare the coefficient of t squared in the product identity. It gives h2=e1 squared minus e2. A repeated variable contributes once to h2 and never to e2, while a product of two distinct variables appears twice in e1 squared and once in each of h2 and e2. This direct coefficient check is useful because it explains the algebraic cancellation rather than treating a basis conversion as an unexplained matrix calculation.
Specializing all N variables to one turns elementary functions into subset counts and complete functions into multiset counts. Their values are binomial coefficients with different upper parameters. Such numerical tests can reject a false identity, but a single all-ones specialization cannot establish equality of symmetric functions: many independent degree-d functions have the same evaluation. Stable symbolic identities retain the individual monomial weights, which later allow tableau contents, representation weights and tensor decompositions to be recovered.
Work it through
Degree two in two visible variables
Specialize to x₁=2 and x₂=3, with every other variable zero. Compute m₂, m₁₁, h₂, e₂ and p₁², then verify their relation.
The exponent pattern (2) gives x₁²+x₂²=4+9=13. Pattern (1,1) gives x₁x₂=6, counted once rather than twice, so m₂=13 and m₁₁=6.
Complete degree two includes both squares and the cross term, hence h₂=19. Elementary degree two permits only distinct variables, hence e₂=6.
The first power sum is 5, so its square is 25. The expression h₂+e₂ also equals 25; symbolically p₁²=m₂+2m₁₁, verifying the different multiplicities.
The values are 13, 6, 19, 6 and 25. The identity h₂+e₂=p₁² survives the specialization with its cross-term multiplicity intact.
The living notebook
Explore the model
Interactive lab · probability
Elementary coefficients become subset counts
Pause, reset, then set success probability p=0.5 and trials n=4. Predict bars 1,4,6,4,1 divided by 16, mean 2 and variance 1. Multiplying each bar by 16 recovers eᵣ(1,1,1,1), the coefficients of (1+t)⁴. The instrument counts independent binary choices; it does not evaluate general symmetric functions or multiset counts.
- 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 forAn all-ones elementary specialization has an exact subset interpretation, while the full symmetric function retains individual weights.
Pause & reason
Check your understanding
Your turn
Try it yourself
Express h₃ in elementary functions and evaluate it on the alphabet (1,1).
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Take the coefficient of t³ in E(−t)H(t)=1; retain h₀=e₀=1.
Compare with the solution
The coefficient equation is h₃−e₁h₂+e₂h₁−e₃=0. Substitute h₁=e₁ and h₂=e₁²−e₂ to obtain h₃=e₁³−2e₁e₂+e₃. At (1,1), e₁=2, e₂=1 and e₃=0, so h₃=8−4=4. The four monomials are x₁³, x₁²x₂, x₁x₂² and x₂³, each evaluating to one.
Keep the idea
Take it with you
- Stable degree-d identities can be checked faithfully in at least d variables.
- Elementary and complete bases are integral; power sums require Q for a basis.
- Evaluation counts objects but forgets part of their weight information.
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