Timeline for how to find explicitly given component in a regular representation
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Mar 25, 2015 at 3:57 | comment | added | Amritanshu Prasad | A function on the vertex-set which sums to $0$ on each face is completely determined by its value at two adjacent vertices. The sum-zero condition will allow you to fix all the other values of this function uniquely and consistently. | |
| Mar 24, 2015 at 17:49 | comment | added | Hari Krovi | I see. Thanks. On the 6 dimensional vertex space, the action of $S_4$ seems to be the same as the one I mentioned. I didn't realize that it can be visualized on an octahedron. It looks like we need only 4 faces to determine a basis for the functions that add up to zero on each face. The other four will be linearly dependent on these. Unless I'm mistaken, it seems to me that the two dimensional representation of $S_4$ is the space orthogonal to this. | |
| Mar 24, 2015 at 16:03 | comment | added | Amritanshu Prasad | You can construct the two dimensional representation of $S_4$ "on the nose" as follows: firstly note that $S_4$ is the group of rigid motions that map the vertex set of a regular (Platonic) octahedron onto itself. Thus $S_4$ also acts on the space of all functions on the vertex-set of this octahedron whose values add up to $0$ on each face. This vector space is the two-dimensional representation of $S_4$. Once you know this, it is easy to write down the matrices in this representation. | |
| Mar 24, 2015 at 15:28 | history | answered | Hari Krovi | CC BY-SA 3.0 |