My professor while writing the character table of $A_{5}$, uses that the span of $S_{5}$ is $\{e_{1} - e_{2}, e_{2} - e_{3}, e_{3} - e_{4}, e_{4} - e_{5} \}$ but I do not know why this is the span of $S_{5}$ or $A_{5}$, could anyone explain this for me please?

  • $\begingroup$ Better ask your professor – I have no idea what the span of a group is, or what all those $e_i$ are. $\endgroup$ – Gerry Myerson Mar 8 at 0:34
  • $\begingroup$ I have asked but I did not understand the answer it was a vague one @GerryMyerson $\endgroup$ – hopefully Mar 8 at 0:49
  • $\begingroup$ @GerryMyerson they are the standard basis $\endgroup$ – hopefully Mar 8 at 0:50
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    $\begingroup$ The standard basis for ${\bf R^5}$, sure, but what does a difference of two vectors in ${\bf R}^5$ have to do with the group $S_5$? If your professor gives you a vague answer, press him/her for a better one. That's what professors get paid for. $\endgroup$ – Gerry Myerson Mar 8 at 1:35

The terminology here isn't correct: It's meaningless to talk about the span of elements of a generic group.

What's probably going on here is this: The group $S_5$ acts faithfully on $\Bbb C^5$ by permutation of coordinates, so that $$\sigma \cdot (z_1, \ldots, z_5) = (z_{\sigma(1)}, \ldots, z_{\sigma(5)}) .$$ This representation is not irreducible: If we denote the standard basis of $\Bbb C^5$ by $(e_1, \ldots, e_5)$, then the action of $S_5$ fixes the line $\{(z, z, z, z, z) : z \in \Bbb C^5\}$, which is spanned by $e_1 + \cdots + e_5$, as well as the complementary hyperplane $\{(z_1, \ldots, z_5) : z_1 + \cdots + z_5 = 0\}$, which is spanned by $e_1 - e_2, \ldots, e_4 - e_5$, and these two subspaces are irreducible subrepresentations.

  • $\begingroup$ is this what is called monomial representation of $S_{n}$? $\endgroup$ – hopefully Mar 8 at 5:24
  • $\begingroup$ so how is the hyperplane is spanned by what you mentioned and why it is the complementary of the line you mentioned? $\endgroup$ – hopefully Mar 8 at 5:32
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    $\begingroup$ The hyperplane has dimension $4$ (it's the kernel of a nontrivial linear map $\Bbb C^5 \to \Bbb C$), and one can check that the $4$ vectors are (1) all in the hyperplane and (2) linearly independent. It it complimentary because (1) the sum of the dimensions of the hyperplane and line is the dimension of the space and the intersection of the hyperplane and line is the trivial vector space. $\endgroup$ – Travis Mar 8 at 5:36
  • $\begingroup$ what about my very first question? $\endgroup$ – hopefully Mar 8 at 5:38
  • $\begingroup$ I don't know offhand. $\endgroup$ – Travis Mar 8 at 5:39

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