Mixing
Motivation or intuition the identity $a=(b-(a-1)^{-1}b(a-1))(a^{-1}ba-(a-1)^{-1}b(a-1))^{-1}$, when $abneq...
3
$begingroup$
In proving the Cartan-Brauer-Hua theorem, Hua uses an obscure identity. That is:
$$
a=(b-(a-1)^{-1}b(a-1))(a^{-1}ba-(a-1)^{-1}b(a-1))^{-1}
$$
for $a$ and $b$ such that $ab neq ba$. All these elements belong to a division ring.
This identity confused me.I want to know some motivation or intuition about this
identity. Thanks for any help.
abstract-algebra ring-theory
asked Jan 28 at 9:47
yuanyuan
393
$endgroup$
|
show 3 more comments
3
$begingroup$
In proving the Cartan-Brauer-Hua theorem, Hua uses an obscure identity. That is:
$$
a=(b-(a-1)^{-1}b(a-1))(a^{-1}ba-(a-1)^{-1}b(a-1))^{-1}
$$
for $a$ and $b$ such that $ab neq ba$. All these elements belong to a division ring.
This identity confused me.I want to know some motivation or intuition about this
identity. Thanks for any help.
abstract-algebra ring-theory
asked Jan 28 at 9:47
yuanyuan
393
$endgroup$
1
$begingroup$
Can you provide some more detail about what kind of objects $a$ and $b$ are? If everything commutes, this is not a well-defined statement. Otherwise it may be but it is still unclear which operations are permitted.
$endgroup$
– quarague
Jan 28 at 10:33
1
$begingroup$
Conjugation in a group is not "obscure".
$endgroup$
– Dietrich Burde
Jan 28 at 10:48
$begingroup$
I think whatever this identity is, it's probably equivalent to Hua's identity? If so, you're re-asking this question of mine, just with a different formulation of the identity.
$endgroup$
– rschwieb
Jan 28 at 12:07
$begingroup$
@rschwieb,does this identity could be implied by Hua's identity?
$endgroup$
– yuan
Jan 28 at 12:13
1
$begingroup$
I have added an answer to this question in math.stackexchange.com/questions/1602573/…
$endgroup$
– Jose Brox
Jan 28 at 16:16
|
show 3 more comments
3
3
3
0
$begingroup$
In proving the Cartan-Brauer-Hua theorem, Hua uses an obscure identity. That is:
$$
a=(b-(a-1)^{-1}b(a-1))(a^{-1}ba-(a-1)^{-1}b(a-1))^{-1}
$$
for $a$ and $b$ such that $ab neq ba$. All these elements belong to a division ring.
This identity confused me.I want to know some motivation or intuition about this
identity. Thanks for any help.
abstract-algebra ring-theory
asked Jan 28 at 9:47
yuanyuan
393
$endgroup$
In proving the Cartan-Brauer-Hua theorem, Hua uses an obscure identity. That is:
$$
a=(b-(a-1)^{-1}b(a-1))(a^{-1}ba-(a-1)^{-1}b(a-1))^{-1}
$$
for $a$ and $b$ such that $ab neq ba$. All these elements belong to a division ring.
This identity confused me.I want to know some motivation or intuition about this
identity. Thanks for any help.
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Jan 28 at 9:47
yuanyuan
393
asked Jan 28 at 9:47
yuanyuan
393
edited Jan 28 at 11:19
yuan
asked Jan 28 at 9:47
yuanyuan
393
asked Jan 28 at 9:47
yuanyuan
393
asked Jan 28 at 9:47
yuanyuan
393
393
1
$begingroup$
Can you provide some more detail about what kind of objects $a$ and $b$ are? If everything commutes, this is not a well-defined statement. Otherwise it may be but it is still unclear which operations are permitted.
$endgroup$
– quarague
Jan 28 at 10:33
1
$begingroup$
Conjugation in a group is not "obscure".
$endgroup$
– Dietrich Burde
Jan 28 at 10:48
$begingroup$
I think whatever this identity is, it's probably equivalent to Hua's identity? If so, you're re-asking this question of mine, just with a different formulation of the identity.
$endgroup$
– rschwieb
Jan 28 at 12:07
$begingroup$
@rschwieb,does this identity could be implied by Hua's identity?
$endgroup$
– yuan
Jan 28 at 12:13
1
$begingroup$
I have added an answer to this question in math.stackexchange.com/questions/1602573/…
$endgroup$
– Jose Brox
Jan 28 at 16:16
|
show 3 more comments
1
$begingroup$
Can you provide some more detail about what kind of objects $a$ and $b$ are? If everything commutes, this is not a well-defined statement. Otherwise it may be but it is still unclear which operations are permitted.
$endgroup$
– quarague
Jan 28 at 10:33
1
$begingroup$
Conjugation in a group is not "obscure".
$endgroup$
– Dietrich Burde
Jan 28 at 10:48
$begingroup$
I think whatever this identity is, it's probably equivalent to Hua's identity? If so, you're re-asking this question of mine, just with a different formulation of the identity.
$endgroup$
– rschwieb
Jan 28 at 12:07
$begingroup$
@rschwieb,does this identity could be implied by Hua's identity?
$endgroup$
– yuan
Jan 28 at 12:13
1
$begingroup$
I have added an answer to this question in math.stackexchange.com/questions/1602573/…
$endgroup$
– Jose Brox
Jan 28 at 16:16
1
1
$begingroup$
Can you provide some more detail about what kind of objects $a$ and $b$ are? If everything commutes, this is not a well-defined statement. Otherwise it may be but it is still unclear which operations are permitted.
$endgroup$
– quarague
Jan 28 at 10:33
$begingroup$
Can you provide some more detail about what kind of objects $a$ and $b$ are? If everything commutes, this is not a well-defined statement. Otherwise it may be but it is still unclear which operations are permitted.
$endgroup$
– quarague
Jan 28 at 10:33
1
1
$begingroup$
Conjugation in a group is not "obscure".
$endgroup$
– Dietrich Burde
Jan 28 at 10:48
$begingroup$
Conjugation in a group is not "obscure".
$endgroup$
– Dietrich Burde
Jan 28 at 10:48
$begingroup$
I think whatever this identity is, it's probably equivalent to Hua's identity? If so, you're re-asking this question of mine, just with a different formulation of the identity.
$endgroup$
– rschwieb
Jan 28 at 12:07
$begingroup$
I think whatever this identity is, it's probably equivalent to Hua's identity? If so, you're re-asking this question of mine, just with a different formulation of the identity.
$endgroup$
– rschwieb
Jan 28 at 12:07
$begingroup$
@rschwieb,does this identity could be implied by Hua's identity?
$endgroup$
– yuan
Jan 28 at 12:13
$begingroup$
@rschwieb,does this identity could be implied by Hua's identity?
$endgroup$
– yuan
Jan 28 at 12:13
1
1
$begingroup$
I have added an answer to this question in math.stackexchange.com/questions/1602573/…
$endgroup$
– Jose Brox
Jan 28 at 16:16
$begingroup$
I have added an answer to this question in math.stackexchange.com/questions/1602573/…
$endgroup$
– Jose Brox
Jan 28 at 16:16
|
show 3 more comments
0
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1
$begingroup$
Can you provide some more detail about what kind of objects $a$ and $b$ are? If everything commutes, this is not a well-defined statement. Otherwise it may be but it is still unclear which operations are permitted.
$endgroup$
– quarague
Jan 28 at 10:33
1
$begingroup$
Conjugation in a group is not "obscure".
$endgroup$
– Dietrich Burde
Jan 28 at 10:48
$begingroup$
I think whatever this identity is, it's probably equivalent to Hua's identity? If so, you're re-asking this question of mine, just with a different formulation of the identity.
$endgroup$
– rschwieb
Jan 28 at 12:07
$begingroup$
@rschwieb,does this identity could be implied by Hua's identity?
$endgroup$
– yuan
Jan 28 at 12:13
1
$begingroup$
I have added an answer to this question in math.stackexchange.com/questions/1602573/…
$endgroup$
– Jose Brox
Jan 28 at 16:16