Mixing
Action of $S_d$ in $V^otimes d$
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
In some books is defined an action of $S_d$ in $V^otimes d$ as $sigma (v_1otimes cdots otimes v_d)=v_sigma^-1(1)otimes cdots otimes v_sigma^-1(d)$ but then
$tau sigma(v_1otimes cdots otimes v_d)=tau (v_sigma^-1(1)otimes cdots otimes v_sigma^-1(d))=v_tau^-1 sigma^-1(1)otimes cdots otimes v_tau^-1 sigma^-1(d)=sigma tau (v_1otimes cdots otimes v_d)$.....
abstract-algebra permutations tensor-products symmetric-groups group-actions
asked Jul 24 at 17:55
Kato
384
add a comment |Â
up vote
2
down vote
favorite
In some books is defined an action of $S_d$ in $V^otimes d$ as $sigma (v_1otimes cdots otimes v_d)=v_sigma^-1(1)otimes cdots otimes v_sigma^-1(d)$ but then
$tau sigma(v_1otimes cdots otimes v_d)=tau (v_sigma^-1(1)otimes cdots otimes v_sigma^-1(d))=v_tau^-1 sigma^-1(1)otimes cdots otimes v_tau^-1 sigma^-1(d)=sigma tau (v_1otimes cdots otimes v_d)$.....
abstract-algebra permutations tensor-products symmetric-groups group-actions
asked Jul 24 at 17:55
Kato
384
1
How about $$tau sigma(v_1otimes cdots otimes v_d)=tau (v_sigma^-1(1)otimes cdots otimes v_sigma^-1(d))=v_sigma^-1tau^-1 (1)otimes cdots otimes v_ sigma^-1tau^-1(d)=tausigma (v_1otimes cdots otimes v_d)?$$
– Lord Shark the Unknown
Jul 24 at 18:03
@Kato if you think about it, the way the action is defined means that whatever is in the i'th place in the pure tensor is transformed into the $sigma(i)$'th place. With this interpretation I think the answer to your inquiry becomes clear.
– Gal Porat
Jul 24 at 18:09
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In some books is defined an action of $S_d$ in $V^otimes d$ as $sigma (v_1otimes cdots otimes v_d)=v_sigma^-1(1)otimes cdots otimes v_sigma^-1(d)$ but then
$tau sigma(v_1otimes cdots otimes v_d)=tau (v_sigma^-1(1)otimes cdots otimes v_sigma^-1(d))=v_tau^-1 sigma^-1(1)otimes cdots otimes v_tau^-1 sigma^-1(d)=sigma tau (v_1otimes cdots otimes v_d)$.....
abstract-algebra permutations tensor-products symmetric-groups group-actions
asked Jul 24 at 17:55
Kato
384
In some books is defined an action of $S_d$ in $V^otimes d$ as $sigma (v_1otimes cdots otimes v_d)=v_sigma^-1(1)otimes cdots otimes v_sigma^-1(d)$ but then
$tau sigma(v_1otimes cdots otimes v_d)=tau (v_sigma^-1(1)otimes cdots otimes v_sigma^-1(d))=v_tau^-1 sigma^-1(1)otimes cdots otimes v_tau^-1 sigma^-1(d)=sigma tau (v_1otimes cdots otimes v_d)$.....
abstract-algebra permutations tensor-products symmetric-groups group-actions
asked Jul 24 at 17:55
Kato
384
asked Jul 24 at 17:55
Kato
384
asked Jul 24 at 17:55
Kato
384
asked Jul 24 at 17:55
Kato
384
384
1
How about $$tau sigma(v_1otimes cdots otimes v_d)=tau (v_sigma^-1(1)otimes cdots otimes v_sigma^-1(d))=v_sigma^-1tau^-1 (1)otimes cdots otimes v_ sigma^-1tau^-1(d)=tausigma (v_1otimes cdots otimes v_d)?$$
– Lord Shark the Unknown
Jul 24 at 18:03
@Kato if you think about it, the way the action is defined means that whatever is in the i'th place in the pure tensor is transformed into the $sigma(i)$'th place. With this interpretation I think the answer to your inquiry becomes clear.
– Gal Porat
Jul 24 at 18:09
add a comment |Â
1
How about $$tau sigma(v_1otimes cdots otimes v_d)=tau (v_sigma^-1(1)otimes cdots otimes v_sigma^-1(d))=v_sigma^-1tau^-1 (1)otimes cdots otimes v_ sigma^-1tau^-1(d)=tausigma (v_1otimes cdots otimes v_d)?$$
– Lord Shark the Unknown
Jul 24 at 18:03
@Kato if you think about it, the way the action is defined means that whatever is in the i'th place in the pure tensor is transformed into the $sigma(i)$'th place. With this interpretation I think the answer to your inquiry becomes clear.
– Gal Porat
Jul 24 at 18:09
1
1
How about $$tau sigma(v_1otimes cdots otimes v_d)=tau (v_sigma^-1(1)otimes cdots otimes v_sigma^-1(d))=v_sigma^-1tau^-1 (1)otimes cdots otimes v_ sigma^-1tau^-1(d)=tausigma (v_1otimes cdots otimes v_d)?$$
– Lord Shark the Unknown
Jul 24 at 18:03
How about $$tau sigma(v_1otimes cdots otimes v_d)=tau (v_sigma^-1(1)otimes cdots otimes v_sigma^-1(d))=v_sigma^-1tau^-1 (1)otimes cdots otimes v_ sigma^-1tau^-1(d)=tausigma (v_1otimes cdots otimes v_d)?$$
– Lord Shark the Unknown
Jul 24 at 18:03
@Kato if you think about it, the way the action is defined means that whatever is in the i'th place in the pure tensor is transformed into the $sigma(i)$'th place. With this interpretation I think the answer to your inquiry becomes clear.
– Gal Porat
Jul 24 at 18:09
@Kato if you think about it, the way the action is defined means that whatever is in the i'th place in the pure tensor is transformed into the $sigma(i)$'th place. With this interpretation I think the answer to your inquiry becomes clear.
– Gal Porat
Jul 24 at 18:09
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
No, that's not how it works.
To understand how $tau$ affects the $v_sigma^-1(1)otimescdotsotimes v_sigma^-1(d)$, you have to represent it in the same form. So, it would be like $w_1otimescdotsotimes w_d$. One possible choice (and the end result does not depend on the choice you make, so, you can take the one you like) would be $w_1=v_sigma^-1(1)$, ..., $w_d=v_sigma^-1(d)$. In short, $w_k=v_sigma^-1(k)$
Then
$$tau(sigma(v_1otimescdotsotimes v_d))=tau(v_sigma^-1otimescdotsotimes v_sigma^-1(d))=tau(w_1otimescdotsotimes w_d)=w_tau^-1(1)otimescdotsotimes w_tau^-1(d)$$
But, since $w_k=v_sigma^-1(k)$, in particular $w_tau^-1(1)=v_sigma^-1(tau^-1(1))$, $w_tau^-1(2)=v_sigma^-1(tau^-1(2))$ and so on. Therefore
$$tau(sigma(v_1otimescdotsotimes v_d))=v_sigma^-1(tau^-1(1))otimescdotsotimes v_sigma^-1(tau^-1(d))=v_(tausigma)^-1(1)otimescdotsotimes v_(tausigma)^-1(d)=(tausigma)(v_1otimescdotsotimes v_d)$$
answered Jul 24 at 18:15
MigMit
27113
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
No, that's not how it works.
To understand how $tau$ affects the $v_sigma^-1(1)otimescdotsotimes v_sigma^-1(d)$, you have to represent it in the same form. So, it would be like $w_1otimescdotsotimes w_d$. One possible choice (and the end result does not depend on the choice you make, so, you can take the one you like) would be $w_1=v_sigma^-1(1)$, ..., $w_d=v_sigma^-1(d)$. In short, $w_k=v_sigma^-1(k)$
Then
$$tau(sigma(v_1otimescdotsotimes v_d))=tau(v_sigma^-1otimescdotsotimes v_sigma^-1(d))=tau(w_1otimescdotsotimes w_d)=w_tau^-1(1)otimescdotsotimes w_tau^-1(d)$$
But, since $w_k=v_sigma^-1(k)$, in particular $w_tau^-1(1)=v_sigma^-1(tau^-1(1))$, $w_tau^-1(2)=v_sigma^-1(tau^-1(2))$ and so on. Therefore
$$tau(sigma(v_1otimescdotsotimes v_d))=v_sigma^-1(tau^-1(1))otimescdotsotimes v_sigma^-1(tau^-1(d))=v_(tausigma)^-1(1)otimescdotsotimes v_(tausigma)^-1(d)=(tausigma)(v_1otimescdotsotimes v_d)$$
answered Jul 24 at 18:15
MigMit
27113
add a comment |Â
up vote
4
down vote
No, that's not how it works.
To understand how $tau$ affects the $v_sigma^-1(1)otimescdotsotimes v_sigma^-1(d)$, you have to represent it in the same form. So, it would be like $w_1otimescdotsotimes w_d$. One possible choice (and the end result does not depend on the choice you make, so, you can take the one you like) would be $w_1=v_sigma^-1(1)$, ..., $w_d=v_sigma^-1(d)$. In short, $w_k=v_sigma^-1(k)$
Then
$$tau(sigma(v_1otimescdotsotimes v_d))=tau(v_sigma^-1otimescdotsotimes v_sigma^-1(d))=tau(w_1otimescdotsotimes w_d)=w_tau^-1(1)otimescdotsotimes w_tau^-1(d)$$
But, since $w_k=v_sigma^-1(k)$, in particular $w_tau^-1(1)=v_sigma^-1(tau^-1(1))$, $w_tau^-1(2)=v_sigma^-1(tau^-1(2))$ and so on. Therefore
$$tau(sigma(v_1otimescdotsotimes v_d))=v_sigma^-1(tau^-1(1))otimescdotsotimes v_sigma^-1(tau^-1(d))=v_(tausigma)^-1(1)otimescdotsotimes v_(tausigma)^-1(d)=(tausigma)(v_1otimescdotsotimes v_d)$$
answered Jul 24 at 18:15
MigMit
27113
add a comment |Â
up vote
4
down vote
up vote
4
down vote
No, that's not how it works.
To understand how $tau$ affects the $v_sigma^-1(1)otimescdotsotimes v_sigma^-1(d)$, you have to represent it in the same form. So, it would be like $w_1otimescdotsotimes w_d$. One possible choice (and the end result does not depend on the choice you make, so, you can take the one you like) would be $w_1=v_sigma^-1(1)$, ..., $w_d=v_sigma^-1(d)$. In short, $w_k=v_sigma^-1(k)$
Then
$$tau(sigma(v_1otimescdotsotimes v_d))=tau(v_sigma^-1otimescdotsotimes v_sigma^-1(d))=tau(w_1otimescdotsotimes w_d)=w_tau^-1(1)otimescdotsotimes w_tau^-1(d)$$
But, since $w_k=v_sigma^-1(k)$, in particular $w_tau^-1(1)=v_sigma^-1(tau^-1(1))$, $w_tau^-1(2)=v_sigma^-1(tau^-1(2))$ and so on. Therefore
$$tau(sigma(v_1otimescdotsotimes v_d))=v_sigma^-1(tau^-1(1))otimescdotsotimes v_sigma^-1(tau^-1(d))=v_(tausigma)^-1(1)otimescdotsotimes v_(tausigma)^-1(d)=(tausigma)(v_1otimescdotsotimes v_d)$$
answered Jul 24 at 18:15
MigMit
27113
No, that's not how it works.
To understand how $tau$ affects the $v_sigma^-1(1)otimescdotsotimes v_sigma^-1(d)$, you have to represent it in the same form. So, it would be like $w_1otimescdotsotimes w_d$. One possible choice (and the end result does not depend on the choice you make, so, you can take the one you like) would be $w_1=v_sigma^-1(1)$, ..., $w_d=v_sigma^-1(d)$. In short, $w_k=v_sigma^-1(k)$
Then
$$tau(sigma(v_1otimescdotsotimes v_d))=tau(v_sigma^-1otimescdotsotimes v_sigma^-1(d))=tau(w_1otimescdotsotimes w_d)=w_tau^-1(1)otimescdotsotimes w_tau^-1(d)$$
But, since $w_k=v_sigma^-1(k)$, in particular $w_tau^-1(1)=v_sigma^-1(tau^-1(1))$, $w_tau^-1(2)=v_sigma^-1(tau^-1(2))$ and so on. Therefore
$$tau(sigma(v_1otimescdotsotimes v_d))=v_sigma^-1(tau^-1(1))otimescdotsotimes v_sigma^-1(tau^-1(d))=v_(tausigma)^-1(1)otimescdotsotimes v_(tausigma)^-1(d)=(tausigma)(v_1otimescdotsotimes v_d)$$
answered Jul 24 at 18:15
MigMit
27113
answered Jul 24 at 18:15
MigMit
27113
answered Jul 24 at 18:15
MigMit
27113
answered Jul 24 at 18:15
MigMit
27113
27113
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861605%2faction-of-s-d-in-v-otimes-d%23new-answer', 'question_page');
);
Post as a guest
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
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
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
1
How about $$tau sigma(v_1otimes cdots otimes v_d)=tau (v_sigma^-1(1)otimes cdots otimes v_sigma^-1(d))=v_sigma^-1tau^-1 (1)otimes cdots otimes v_ sigma^-1tau^-1(d)=tausigma (v_1otimes cdots otimes v_d)?$$
– Lord Shark the Unknown
Jul 24 at 18:03
@Kato if you think about it, the way the action is defined means that whatever is in the i'th place in the pure tensor is transformed into the $sigma(i)$'th place. With this interpretation I think the answer to your inquiry becomes clear.
– Gal Porat
Jul 24 at 18:09