The monomial representation of $S_n$ is irreducible.

Question

I am going through "Linear group representations" by Ernest Vinberg The example is given below:

But I have difficulties in understanding the following:

1- why $V_{0}$ is called $(n-1)-$dimensional subspace, I want a concrete example please?

2- Why if the characteristic of the field $K$ is equal to zero, then $V_{1}$ is not subset of $V_{0}$?

3- I do not understand how he calculated $M((12))x - x$, could anyone clarify this for me?

4- Now I am stuck on this question (the above example is example 5):

Answer 1

I can answer to the first three questions:

• $V_0$ is defined by a single linear equation. The general result is this: if a subspace is defined by a system of linear equations of rank $r$ (i.e. the matrix with rows the coefficients of the defining linear equations has rank $r$), this subspace has codimension $r$, which means that in a space of dimension $n$, it has dimension $n-r$. This is because you can determine $r$ coordinates in function of the $n-r$ others (cf. the way to solve linear systems of equations).
• $V_0$ is defined by the equation $x_1+x_2+ +\dots +x_n=0$, whereas all vectors in $V_1$ satisfy $x_i=x_j \:\forall i,j$, so that for such a vector, $x_1+x_2+\dots+x_n=nx_1$, which is $\ne 0$, except for the null vector, or if the characteristic of $K$ is a prime factor of $n$.
• If $x=x_1e_1+x_2e_2+\sum_{i=3}^n x_ie_i$, we have, by linearity, $$M((1\,2))x=x_1e_2+x_2e_1+\sum_{i=3}^n x_ie_i,$$ so that \begin{align} M((1\,2))x-x&=x_1e_2+x_2e_1+\sum_{i=3}^n x_ie_i-\Bigl(x_1e_1+x_2e_2+\sum_{i=3}^n x_ie_i\Bigr),\\ &=x_1e_2+x_2e_1-(x_1e_1+x_2e_2)=x_2e_1-x_1e_1+x_1e_2-x_2e_2\\ &=\dotsm \end{align}

Answer 2

1. It is $(n-1)$-dimensional because it has a basis with $n-1$ elements:$$\{e_1-e_2,e_2-e_3,\ldots,e_{n-1}-e_n\}.$$
2. The set $V_1$ is not a subset of $V_0$ because $e_1+e_2+\cdots+e_n\in V_1\setminus V_0$, since $\overbrace{1+1+\cdots+1}^{n\text{ terms}}\neq0$ (because $K$ has characteristic $0$).
3. Did you read the definition if $M(\sigma)$? By this definition, $M\bigl((1\ \ 2)\bigr)(e_1)=e_2$, $M\bigl((1\ \ 2)\bigr)(e_2)=e_1$, and $M\bigl((1\ \ 2)\bigr)(e_k)=e_k$ if $k>2$. Therefore, if $x=x_1e_1+x_2e_2+\cdots+x_ne_n$, then$$M\bigl((1\ \ 2)\bigr)(x_1e_1+x_2e_2+\cdots+x_ne_n)=x_2e_1+x_1e_2+x_3e_3+\cdots+x_ne_n.$$
4. I suspect that Vinberg stated some result (about group characters, perhaps) to prove this, but I don't have his book at hand right now.