Mixing
If every group element has a fixed vector, does there exist a vector fixed by all?
$begingroup$
Let $G leq SL(V)$ be a finite group where $V$ is a finite dimensional linear space over a finite field $k$, and assume that every element in $G$ has a nonzero fixed point in $V$, then is it necessary that there exist a nonzero point fixed by $G$ ?
I think in the case $dim V=2$, this is true. For example, if $G leq SL_2(k)$, then one consider the fixed line $l_g$ for any $g not =1 in G$, then $l_g$ must be the same or one choose two lines as a basis then get a contradiction by considering the product.
abstract-algebra group-theory representation-theory
asked Jan 29 at 23:13
zzyzzy
2,6431420
$endgroup$
add a comment |
$begingroup$
Let $G leq SL(V)$ be a finite group where $V$ is a finite dimensional linear space over a finite field $k$, and assume that every element in $G$ has a nonzero fixed point in $V$, then is it necessary that there exist a nonzero point fixed by $G$ ?
I think in the case $dim V=2$, this is true. For example, if $G leq SL_2(k)$, then one consider the fixed line $l_g$ for any $g not =1 in G$, then $l_g$ must be the same or one choose two lines as a basis then get a contradiction by considering the product.
abstract-algebra group-theory representation-theory
asked Jan 29 at 23:13
zzyzzy
2,6431420
$endgroup$
add a comment |
$begingroup$
Let $G leq SL(V)$ be a finite group where $V$ is a finite dimensional linear space over a finite field $k$, and assume that every element in $G$ has a nonzero fixed point in $V$, then is it necessary that there exist a nonzero point fixed by $G$ ?
I think in the case $dim V=2$, this is true. For example, if $G leq SL_2(k)$, then one consider the fixed line $l_g$ for any $g not =1 in G$, then $l_g$ must be the same or one choose two lines as a basis then get a contradiction by considering the product.
abstract-algebra group-theory representation-theory
asked Jan 29 at 23:13
zzyzzy
2,6431420
$endgroup$
Let $G leq SL(V)$ be a finite group where $V$ is a finite dimensional linear space over a finite field $k$, and assume that every element in $G$ has a nonzero fixed point in $V$, then is it necessary that there exist a nonzero point fixed by $G$ ?
I think in the case $dim V=2$, this is true. For example, if $G leq SL_2(k)$, then one consider the fixed line $l_g$ for any $g not =1 in G$, then $l_g$ must be the same or one choose two lines as a basis then get a contradiction by considering the product.
abstract-algebra group-theory representation-theory
abstract-algebra group-theory representation-theory
asked Jan 29 at 23:13
zzyzzy
2,6431420
asked Jan 29 at 23:13
zzyzzy
2,6431420
edited Jan 29 at 23:44
zzy
asked Jan 29 at 23:13
zzyzzy
2,6431420
asked Jan 29 at 23:13
zzyzzy
2,6431420
asked Jan 29 at 23:13
zzyzzy
2,6431420
2,6431420
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No. For instance, suppose $operatorname{char} kneq 2$, let $V=k^3$, and let $Gsubseteq SL(V)$ be the subgroup generated by $$begin{pmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0end{pmatrix}$$ and $$begin{pmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1end{pmatrix}.$$ Explicitly, an element of $G$ is an even permutation matrix except that two of its $1$s may be changed to $-1$. Every element of $G$ fixes some nonzero vector: if the permutation is the identity then some standard basis vector is fixed, if the permutation is a 3-cycle and there are no $-1$s then $(1,1,1)$ is fixed, and if the permutation is a 3-cycle and there are two $-1$s then some permutation of the vector $(1,1,-1)$ is fixed. However, no nonzero vector is fixed by $G$ (the first generator only fixes scalar multiples of $(1,1,1)$, which are not fixed by the second generator).
answered Jan 30 at 0:49
Eric WofseyEric Wofsey
192k14216350
$endgroup$
1
$begingroup$
I was approaching a very similar example, by starting with the canonical representation $SO(3)$ on $mathbb{R}^3$ (in which every rotation has its axis as a fixed point), then taking a finite subgroup of $SO(3)$ such as the matrices which fix the union of the coordinate axes, then replacing $mathbb{R}$ with a finite field with sufficient characteristic constraints and sufficient existence of roots of certain polynomials to enable the example still to work for that finite field.
$endgroup$
– Daniel Schepler
Jan 30 at 1:11
$begingroup$
@DanielSchepler: That was my inspiration too! I was messing around with trying to show every element of $SO(3)$ over $mathbb{F}_3$ has a fixed point when I realized I could simplify things by just looking at this subgroup, which would work over any characteristic except $2$.
$endgroup$
– Eric Wofsey
Jan 30 at 1:20
$begingroup$
Thank you for the good answer !
$endgroup$
– zzy
Jan 30 at 1:34
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092862%2fif-every-group-element-has-a-fixed-vector-does-there-exist-a-vector-fixed-by-al%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. For instance, suppose $operatorname{char} kneq 2$, let $V=k^3$, and let $Gsubseteq SL(V)$ be the subgroup generated by $$begin{pmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0end{pmatrix}$$ and $$begin{pmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1end{pmatrix}.$$ Explicitly, an element of $G$ is an even permutation matrix except that two of its $1$s may be changed to $-1$. Every element of $G$ fixes some nonzero vector: if the permutation is the identity then some standard basis vector is fixed, if the permutation is a 3-cycle and there are no $-1$s then $(1,1,1)$ is fixed, and if the permutation is a 3-cycle and there are two $-1$s then some permutation of the vector $(1,1,-1)$ is fixed. However, no nonzero vector is fixed by $G$ (the first generator only fixes scalar multiples of $(1,1,1)$, which are not fixed by the second generator).
answered Jan 30 at 0:49
Eric WofseyEric Wofsey
192k14216350
$endgroup$
1
$begingroup$
I was approaching a very similar example, by starting with the canonical representation $SO(3)$ on $mathbb{R}^3$ (in which every rotation has its axis as a fixed point), then taking a finite subgroup of $SO(3)$ such as the matrices which fix the union of the coordinate axes, then replacing $mathbb{R}$ with a finite field with sufficient characteristic constraints and sufficient existence of roots of certain polynomials to enable the example still to work for that finite field.
$endgroup$
– Daniel Schepler
Jan 30 at 1:11
$begingroup$
@DanielSchepler: That was my inspiration too! I was messing around with trying to show every element of $SO(3)$ over $mathbb{F}_3$ has a fixed point when I realized I could simplify things by just looking at this subgroup, which would work over any characteristic except $2$.
$endgroup$
– Eric Wofsey
Jan 30 at 1:20
$begingroup$
Thank you for the good answer !
$endgroup$
– zzy
Jan 30 at 1:34
add a comment |
$begingroup$
No. For instance, suppose $operatorname{char} kneq 2$, let $V=k^3$, and let $Gsubseteq SL(V)$ be the subgroup generated by $$begin{pmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0end{pmatrix}$$ and $$begin{pmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1end{pmatrix}.$$ Explicitly, an element of $G$ is an even permutation matrix except that two of its $1$s may be changed to $-1$. Every element of $G$ fixes some nonzero vector: if the permutation is the identity then some standard basis vector is fixed, if the permutation is a 3-cycle and there are no $-1$s then $(1,1,1)$ is fixed, and if the permutation is a 3-cycle and there are two $-1$s then some permutation of the vector $(1,1,-1)$ is fixed. However, no nonzero vector is fixed by $G$ (the first generator only fixes scalar multiples of $(1,1,1)$, which are not fixed by the second generator).
answered Jan 30 at 0:49
Eric WofseyEric Wofsey
192k14216350
$endgroup$
1
$begingroup$
I was approaching a very similar example, by starting with the canonical representation $SO(3)$ on $mathbb{R}^3$ (in which every rotation has its axis as a fixed point), then taking a finite subgroup of $SO(3)$ such as the matrices which fix the union of the coordinate axes, then replacing $mathbb{R}$ with a finite field with sufficient characteristic constraints and sufficient existence of roots of certain polynomials to enable the example still to work for that finite field.
$endgroup$
– Daniel Schepler
Jan 30 at 1:11
$begingroup$
@DanielSchepler: That was my inspiration too! I was messing around with trying to show every element of $SO(3)$ over $mathbb{F}_3$ has a fixed point when I realized I could simplify things by just looking at this subgroup, which would work over any characteristic except $2$.
$endgroup$
– Eric Wofsey
Jan 30 at 1:20
$begingroup$
Thank you for the good answer !
$endgroup$
– zzy
Jan 30 at 1:34
add a comment |
$begingroup$
No. For instance, suppose $operatorname{char} kneq 2$, let $V=k^3$, and let $Gsubseteq SL(V)$ be the subgroup generated by $$begin{pmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0end{pmatrix}$$ and $$begin{pmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1end{pmatrix}.$$ Explicitly, an element of $G$ is an even permutation matrix except that two of its $1$s may be changed to $-1$. Every element of $G$ fixes some nonzero vector: if the permutation is the identity then some standard basis vector is fixed, if the permutation is a 3-cycle and there are no $-1$s then $(1,1,1)$ is fixed, and if the permutation is a 3-cycle and there are two $-1$s then some permutation of the vector $(1,1,-1)$ is fixed. However, no nonzero vector is fixed by $G$ (the first generator only fixes scalar multiples of $(1,1,1)$, which are not fixed by the second generator).
answered Jan 30 at 0:49
Eric WofseyEric Wofsey
192k14216350
$endgroup$
No. For instance, suppose $operatorname{char} kneq 2$, let $V=k^3$, and let $Gsubseteq SL(V)$ be the subgroup generated by $$begin{pmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0end{pmatrix}$$ and $$begin{pmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1end{pmatrix}.$$ Explicitly, an element of $G$ is an even permutation matrix except that two of its $1$s may be changed to $-1$. Every element of $G$ fixes some nonzero vector: if the permutation is the identity then some standard basis vector is fixed, if the permutation is a 3-cycle and there are no $-1$s then $(1,1,1)$ is fixed, and if the permutation is a 3-cycle and there are two $-1$s then some permutation of the vector $(1,1,-1)$ is fixed. However, no nonzero vector is fixed by $G$ (the first generator only fixes scalar multiples of $(1,1,1)$, which are not fixed by the second generator).
answered Jan 30 at 0:49
Eric WofseyEric Wofsey
192k14216350
answered Jan 30 at 0:49
Eric WofseyEric Wofsey
192k14216350
answered Jan 30 at 0:49
Eric WofseyEric Wofsey
192k14216350
answered Jan 30 at 0:49
Eric WofseyEric Wofsey
192k14216350
192k14216350
1
$begingroup$
I was approaching a very similar example, by starting with the canonical representation $SO(3)$ on $mathbb{R}^3$ (in which every rotation has its axis as a fixed point), then taking a finite subgroup of $SO(3)$ such as the matrices which fix the union of the coordinate axes, then replacing $mathbb{R}$ with a finite field with sufficient characteristic constraints and sufficient existence of roots of certain polynomials to enable the example still to work for that finite field.
$endgroup$
– Daniel Schepler
Jan 30 at 1:11
$begingroup$
@DanielSchepler: That was my inspiration too! I was messing around with trying to show every element of $SO(3)$ over $mathbb{F}_3$ has a fixed point when I realized I could simplify things by just looking at this subgroup, which would work over any characteristic except $2$.
$endgroup$
– Eric Wofsey
Jan 30 at 1:20
$begingroup$
Thank you for the good answer !
$endgroup$
– zzy
Jan 30 at 1:34
add a comment |
1
$begingroup$
I was approaching a very similar example, by starting with the canonical representation $SO(3)$ on $mathbb{R}^3$ (in which every rotation has its axis as a fixed point), then taking a finite subgroup of $SO(3)$ such as the matrices which fix the union of the coordinate axes, then replacing $mathbb{R}$ with a finite field with sufficient characteristic constraints and sufficient existence of roots of certain polynomials to enable the example still to work for that finite field.
$endgroup$
– Daniel Schepler
Jan 30 at 1:11
$begingroup$
@DanielSchepler: That was my inspiration too! I was messing around with trying to show every element of $SO(3)$ over $mathbb{F}_3$ has a fixed point when I realized I could simplify things by just looking at this subgroup, which would work over any characteristic except $2$.
$endgroup$
– Eric Wofsey
Jan 30 at 1:20
$begingroup$
Thank you for the good answer !
$endgroup$
– zzy
Jan 30 at 1:34
1
1
$begingroup$
I was approaching a very similar example, by starting with the canonical representation $SO(3)$ on $mathbb{R}^3$ (in which every rotation has its axis as a fixed point), then taking a finite subgroup of $SO(3)$ such as the matrices which fix the union of the coordinate axes, then replacing $mathbb{R}$ with a finite field with sufficient characteristic constraints and sufficient existence of roots of certain polynomials to enable the example still to work for that finite field.
$endgroup$
– Daniel Schepler
Jan 30 at 1:11
$begingroup$
I was approaching a very similar example, by starting with the canonical representation $SO(3)$ on $mathbb{R}^3$ (in which every rotation has its axis as a fixed point), then taking a finite subgroup of $SO(3)$ such as the matrices which fix the union of the coordinate axes, then replacing $mathbb{R}$ with a finite field with sufficient characteristic constraints and sufficient existence of roots of certain polynomials to enable the example still to work for that finite field.
$endgroup$
– Daniel Schepler
Jan 30 at 1:11
$begingroup$
@DanielSchepler: That was my inspiration too! I was messing around with trying to show every element of $SO(3)$ over $mathbb{F}_3$ has a fixed point when I realized I could simplify things by just looking at this subgroup, which would work over any characteristic except $2$.
$endgroup$
– Eric Wofsey
Jan 30 at 1:20
$begingroup$
@DanielSchepler: That was my inspiration too! I was messing around with trying to show every element of $SO(3)$ over $mathbb{F}_3$ has a fixed point when I realized I could simplify things by just looking at this subgroup, which would work over any characteristic except $2$.
$endgroup$
– Eric Wofsey
Jan 30 at 1:20
$begingroup$
Thank you for the good answer !
$endgroup$
– zzy
Jan 30 at 1:34
$begingroup$
Thank you for the good answer !
$endgroup$
– zzy
Jan 30 at 1:34
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092862%2fif-every-group-element-has-a-fixed-vector-does-there-exist-a-vector-fixed-by-al%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
