Timeline for showing that $G$ does not have an embedding in $GL_n(F)$ for any $n\ge 1$ and field $F$
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 4, 2021 at 19:31 | comment | added | Jyrki Lahtonen | Related. | |
| Aug 23, 2021 at 14:42 | vote | accept | Alfred | ||
| Aug 19, 2021 at 4:12 | comment | added | Jyrki Lahtonen | (cont'd) In the end (the end must come because $V$ is finite dimensional) no further splitting into lower dimensional subspaces is possible, and by respect the decomposition we have shared eigenspaces. | |
| Aug 19, 2021 at 4:11 | comment | added | Jyrki Lahtonen | It may be that calling the argument induction on $k$ and $\dim V$ may be a stretch. It is possible to organize the argument differently. First split $V$ into a direct sum of two subspace that are the eigenspaces of $\phi(\sigma_1)$. Then respect the decomposition allows us to split both $V_+$ and $V_-$ into direct sums of eigenspaces of $\phi(\sigma_2)$. At this point we have a direct sum into four components (some possibly trivial). Again respect the decomposition shows that all four are stable under $\phi(\sigma_3)$, and we can split each in two. Et cetera. | |
| Aug 19, 2021 at 4:05 | comment | added | Jyrki Lahtonen | Alfred, are you familiar with extension fields? Pieces of that are needed when we move from $F$ to $F(\omega)$. | |
| Aug 19, 2021 at 4:04 | comment | added | Jyrki Lahtonen | Inductive step: If $V_+=V$ or $V_-=V$ there is nothing to do at the induction step because $\phi(\sigma_{k+1})$ is $\pm id_V$. In other cases $\dim V_+<n$ and $\dim V_-<n$, so we can use induction hypothesis on $n$ and the part that that all the transformations $\phi(\sigma_j)$, $j<k+1$, respect the decomposition to conclude that their restrictions to both $V_+$ and $V_-$ have shared eigenbases. Putting these eigenbases together gives an eigenbasis for all of $V$ as claimed. | |
| Aug 19, 2021 at 3:59 | comment | added | Jyrki Lahtonen | If $\omega\in F$ there is no need to use anything but $F$. Otherwise we replace $F$ with the extension field $F(\omega)$. By elementary theory of field extensions $[F(\omega):F]=2$ so $F(\omega)=\{a+b\omega\mid a,b\in F\}$ as $\omega^2+\omega+1=0$. $GL_n(F)$ is a subgroup of $GL_n(F(\omega)$, so if it is impossible to embed $G$ into $GL_n(F(\omega))$ it is impossible to embed it into $GL_n(F)$ either. | |
| Aug 19, 2021 at 3:55 | comment | added | Jyrki Lahtonen | We have shown that with respect to a suitable basis the matrices $\phi(\sigma_j)$, $j=1,2,\ldots,2^n+1$ are all diagonal with entries $\pm1$. As there are only $2^n$ such matrices we must have $\phi(\sigma_{j_1})=\phi(\sigma_{j_2})$ for some $j_1\neq j_2$. This violates the assumption that $\phi$ is an embedding. | |
| Aug 19, 2021 at 3:54 | comment | added | Alfred | Could you elaborate on how you proved your inductive step when you assumed $char(F)\neq 2$? | |
| Aug 19, 2021 at 3:53 | comment | added | Jyrki Lahtonen | A vector $x$ is a shared eigenvector of linear transformations $T_1,\ldots,T_\ell$ when $T_j(x)=\lambda_j(x)$ for all $j$. Observe that the eigenvalue $\lambda_j$ may depend on $j$. | |
| Aug 19, 2021 at 3:51 | comment | added | Jyrki Lahtonen | They respect the decomposition means exactly what you wrote above: $\phi(\sigma_j)(V_+)=V_+$ and $\phi(\sigma_j)(V_-)=V_-$. | |
| Aug 18, 2021 at 21:26 | history | edited | Alfred | CC BY-SA 4.0 | added 8 characters in body |
| Aug 18, 2021 at 21:12 | history | edited | Alfred | CC BY-SA 4.0 | added 4053 characters in body |
| Aug 18, 2021 at 14:06 | history | edited | Alfred | CC BY-SA 4.0 | deleted 8 characters in body |
| Aug 18, 2021 at 4:48 | answer | added | Jyrki Lahtonen | timeline score: 4 | |
| Aug 18, 2021 at 4:41 | answer | added | Eric Wofsey | timeline score: 5 | |
| Aug 18, 2021 at 4:04 | comment | added | Jyrki Lahtonen | If $1+1\neq0$ in $F$ then we could use the fact $G$ contains infinitely many commuting 2-cycles. Their images would need to be simultaneously diagonalizable (by a bit of linear algebra), but there's no room on the diagonal. In characteristic two we could adjoin a thrid root of unity to $F$ and use disjoint 3-cycles to reach the same conclusion. | |
| Aug 18, 2021 at 3:45 | history | asked | Alfred | CC BY-SA 4.0 |