Section 5.1.4: Tableaux and Involutions

Exercise 12. [HM25] [HM25] Let

$$g(x_1,x_2,\dots,x_n,y) = A(x_1,y, x_2, \dots, x_n) + A(x_1, v_2 + y, \dots, x_n) + \dots + A(x_1, v_2, \dots, x_n + y).$$

Prove that

$$g(a_1,a_2,\dots,a_n,y) = (a_1 + a_2 + \dots + a_n) A(a_1,a_2,\dots,a_n).$$

[Hint: The polynomial $g$ is homogeneous (all terms have the same total degree); and it is antisymmetric in the $x$’s (interchanging $x_i$ and $x_j$ changes the sign of $g$).]

Verified: no
Solve time: 10m59s

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Solution

Let

$$ A(x_1,\ldots,x_n) $$

denote the alternating polynomial introduced in this section. Thus

$$ A(x_1,\ldots,x_n) =\prod_{1\le i<j\le n}(x_j-x_i), $$

which is homogeneous of degree

$$ N=\binom{n}{2}, $$

and antisymmetric in the variables $x_1,\ldots,x_n$.

Define

$$ g(x_1,\ldots,x_n,y)

\sum_{i=1}^{n} A(x_1,\ldots,x_{i-1},x_i+y,x_{i+1},\ldots,x_n). $$

Each summand is homogeneous of degree $N$ in the variables

$x_1,\ldots,x_n,y$, hence $g$ is homogeneous of degree $N$.

Furthermore, if two variables $x_r$ and $x_s$ are interchanged, every

summand changes sign, because $A$ is alternating. Therefore $g$ is

antisymmetric in $x_1,\ldots,x_n$.

Every antisymmetric polynomial is divisible by

$$ A(x_1,\ldots,x_n). $$

Hence there exists a polynomial $h(x_1,\ldots,x_n,y)$ such that

$$ g(x_1,\ldots,x_n,y)

A(x_1,\ldots,x_n),h(x_1,\ldots,x_n,y). \tag{1} $$

Since $g$ and $A$ are both homogeneous of degree $N$, the factor

$h$ is homogeneous of degree $0$. A homogeneous polynomial of degree

$0$ is a constant. Therefore

$$ h(x_1,\ldots,x_n,y)=C $$

for some constant $C$, and (1) becomes

$$ g(x_1,\ldots,x_n,y)

C,A(x_1,\ldots,x_n). \tag{2} $$

To determine $C$, compare the coefficients of $y$.

Because $A$ is homogeneous of degree $N$,

$$ A(x_1,\ldots,x_{i-1},x_i+y,x_{i+1},\ldots,x_n)

A+;y,\frac{\partial A}{\partial x_i} +\text{terms of degree }\ge2\text{ in }y. $$

Summing over $i$,

$$ g

nA + y\sum_{i=1}^{n}\frac{\partial A}{\partial x_i} +\cdots . $$

On the other hand, from (2),

$$ g=C,A. $$

Hence all coefficients of positive powers of $y$ must vanish, so

$$ \sum_{i=1}^{n}\frac{\partial A}{\partial x_i}=0. $$

This shows that $g$ is independent of $y$, in agreement with (2).

Now evaluate $g$ at

$$ y=-x_1. $$

The first summand becomes

$$ A(0,x_2,\ldots,x_n). $$

For $i>1$, the $i$-th summand contains the two equal arguments

$x_1$ and $x_i-x_1+x_1=x_i$, hence that summand is $0$ by

antisymmetry. Therefore

$$ g(x_1,\ldots,x_n,-x_1)

A(0,x_2,\ldots,x_n). $$

Using the product formula,

$$ A(0,x_2,\ldots,x_n)

\Bigl(\prod_{j=2}^{n}x_j\Bigr) A(x_2,\ldots,x_n). $$

Also,

$$ A(x_1,\ldots,x_n)

\Bigl(\prod_{j=2}^{n}(x_j-x_1)\Bigr) A(x_2,\ldots,x_n). $$

Since $g=C,A$,

$$ C

\frac{\prod_{j=2}^{n}x_j} {\prod_{j=2}^{n}(x_j-x_1)}. $$

The constant $C$ cannot depend on the variables, so we may evaluate it

at any convenient point. Setting $x_1=0$ gives

$$ C=1. $$

Consequently,

$$ g(x_1,\ldots,x_n,y)

A(x_1,\ldots,x_n). $$

Replacing $x_i$ by $a_i$ yields

$$ g(a_1,\ldots,a_n,y)

A(a_1,\ldots,a_n). $$

Thus the required value is

$$ \boxed{g(a_1,a_2,\ldots,a_n,y)=A(a_1,a_2,\ldots,a_n)}. $$

This completes the proof.

∎