In answers to some of my previous questions , a lot of people used the terms homogeneous polynomial ( in a,b,c ) (under permutations of variables ) , cyclic polynomial ( in a,b,c) (under permutations of variables ) , anti-symmetric and symmetric polynomial . Please explain their meaning .
Also in answering my previous question , Barry Cipra said , ( link )

" The expression $P(a,b,c)=(a+b+c)(−a+b+c)(a−b+c)(a+b−c)$ is clearly a homogeneous polynomial of degree 4, symmetric in its three variables. It's also clear that the coefficient of $a^4$ (hence also $b^4$ and $c^4$) is −1. Moreover, $P(−a,b,c)=P(a,−b,c)=P(a,b,−c)=P(a,b,c)$ which implies P has no terms with any variable taken to an odd degree."

I don't understand why P has no variable with odd degree. So please explain that too.

( For context , see my previous questions :
Simplify : $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7) $
Factorize: $a^2(b − c)^3 + b^2(c − a)^3 + c^2(a − b)^3$
Factorise: $2a^4 + a^2b^2 + ab^3 + b^4$ )

  • $\begingroup$ Homogenous polynomial :IF Each term in a polynomial has same degree,it is called a homogenous polynomial.The expression x^2+2xy+5y^2 is homogenous of variable x and y with degree two .It is a special case of the homogenous polynomial ax^2+2axy+by^2 $\endgroup$
    – rah4927
    Aug 11, 2013 at 18:03
  • $\begingroup$ Another way of seeing homogeneous polynomials is the following: Suppose $X=(x_1,\cdots, x_n)$ then $P(X)$ is homogeneous of order $k$ if $P(tX)=t^k P(X)$ for any scalar $t$. $\endgroup$
    – Maesumi
    Aug 11, 2013 at 18:31
  • $\begingroup$ If $P(X)=P(\sigma (X))$ for any permutation $\sigma$ then $P$ is symmetric. $\endgroup$
    – Maesumi
    Aug 11, 2013 at 19:38
  • $\begingroup$ See definition of parity. If $P(X)= (-1)^m P(\sigma (X))$ for any permutation $\sigma$ of parity $m$ then $P$ is anti-symmetric. $\endgroup$
    – Maesumi
    Aug 11, 2013 at 19:40
  • $\begingroup$ If $P(X)=P(\sigma (X))$ for any cyclic permutation $\sigma$ then $P$ is cyclic. $\endgroup$
    – Maesumi
    Aug 11, 2013 at 19:42

1 Answer 1



A function $f$ of $n$ variables is homogeneous of degree $k$ if

$$ f(tx_1, \dots, tx_n) = t^k f(x_1, \dots, x_n). $$

For example:

$$ x^2 + y^2,\quad xyz,\quad x^4 - 5x^3y,\quad \cos \frac{x}{y}. $$

Note that the total degree of each term of the polynomials are the same. For $x^4-5x^3y$, the total degree of the first term is $4 + 0 = 4$, and for the second term it's $3 + 1 = 4$.

Euler showed that this implies for differentiable $f$:

$$ x_1 \frac{\partial f}{\partial x_1} + \cdots + x_n \frac{\partial f}{\partial x_n} = k f. $$


A function f is symmetric if

$$ f(\dots, x_i, \dots, x_j, \dots) = f(\dots, x_j, \dots, x_i, \dots)\quad\text{for all }i, j. $$

For example,

$$ x + y,\quad xyz,\quad xy + xz + yz,\quad x^8 + y^8 + z^8 + w^8. $$ If a symmetric polynomial has a monomial $x^3 y^2 z$, it also has a monomial $x y^3 z^2$, for example. Any symmetric polynomial is a polynomial in the elementary symmetric functions:

$$ \begin{align} e_0(x_1, \dots, x_n) &= 1 \\ e_1(x_1, \dots, x_n) &=\sum_{1\le i \le n} x_i \\ e_2(x_1, \dots, x_n) &=\sum_{1\le i < j \le n} x_i x_j\\ \dots\\ e_n(x_1, \dots, x_n) &= x_1 x_2 \cdots x_n. \end{align} $$


A function f is anti-symmetric if

$$ f(\dots, x_i, \dots, x_j, \dots) = - f(\dots, x_j, \dots, x_i, \dots)\quad\text{for all }i, j. $$

For example,

$$ (x-y) (x-z) (y-z) $$ What happens to this if you swap $x$ and $y$? $$ (y-x)(y-z)(x-z) = -(x-y) (x-z) (y-z) $$ Swapping $x$ and $z$, or swapping $y$ and $z$ does the same.


A polynomial is cyclic if

$$ f(x_1, \dots, x_n) = f(x_2, \ldots, x_n, x_1) $$

This implies that

$$ f(x_1, \dots, x_n) = f(x_{k+1}, \ldots, x_n, x_1, \dots, x_k)\quad\mbox{for any } k. $$

For example, $x^3y + y^3z + z^3x$. If you replace $(x, y, z)$ with $(y, z, x)$, you get $y^3z + z^3x + x^3 y$, which equals the original.

  • $\begingroup$ Can you please give example polynomials for each definition ? I particularly don't understand the homogeneous definition. $\endgroup$
    – A Googler
    Oct 1, 2013 at 6:25

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