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Question about the $operatornameExp$ map on $operatornameEnd(V)$ and the left invariant vector field

2

$begingroup$

Let $V$ be a finite dimensional vector space over $mathbbR$.
We define for $T in operatornameEnd(V)$ with $| T | < infty$
$$
operatornameExp(T) = sum_k=0^infty fracT^kk!.
$$
Then it can be shown that
$$
fracddt operatornameExp(tT)= operatornameExp(tT)cdot T.
$$

Then the notes I am reading states :
"Therefore, $t rightarrow operatornameExp(tT)$ is tangential to the left-invariant vector field evaluated at $e$ is $T$."

I am struggling to understand this sentence here. I would appreciate if someone could explain me in more detail what this sentence means... In particular, which left-invariant vector field are they talking about?
Thank you.

linear-algebra differential-geometry vector-spaces lie-groups lie-algebras

share|cite|improve this question

edited Apr 1 at 17:42

Bernard

124k742117

asked Apr 1 at 16:33

Takeshi GoudaTakeshi Gouda

645

$endgroup$

add a comment | 

2

$begingroup$

Let $V$ be a finite dimensional vector space over $mathbbR$.
We define for $T in operatornameEnd(V)$ with $| T | < infty$
$$
operatornameExp(T) = sum_k=0^infty fracT^kk!.
$$
Then it can be shown that
$$
fracddt operatornameExp(tT)= operatornameExp(tT)cdot T.
$$

Then the notes I am reading states :
"Therefore, $t rightarrow operatornameExp(tT)$ is tangential to the left-invariant vector field evaluated at $e$ is $T$."

I am struggling to understand this sentence here. I would appreciate if someone could explain me in more detail what this sentence means... In particular, which left-invariant vector field are they talking about?
Thank you.

linear-algebra differential-geometry vector-spaces lie-groups lie-algebras

share|cite|improve this question

edited Apr 1 at 17:42

Bernard

124k742117

asked Apr 1 at 16:33

Takeshi GoudaTakeshi Gouda

645

$endgroup$

add a comment | 

$begingroup$

Let $V$ be a finite dimensional vector space over $mathbbR$.
We define for $T in operatornameEnd(V)$ with $| T | < infty$
$$
operatornameExp(T) = sum_k=0^infty fracT^kk!.
$$
Then it can be shown that
$$
fracddt operatornameExp(tT)= operatornameExp(tT)cdot T.
$$

Then the notes I am reading states :
"Therefore, $t rightarrow operatornameExp(tT)$ is tangential to the left-invariant vector field evaluated at $e$ is $T$."

I am struggling to understand this sentence here. I would appreciate if someone could explain me in more detail what this sentence means... In particular, which left-invariant vector field are they talking about?
Thank you.

linear-algebra differential-geometry vector-spaces lie-groups lie-algebras

share|cite|improve this question

edited Apr 1 at 17:42

Bernard

124k742117

asked Apr 1 at 16:33

Takeshi GoudaTakeshi Gouda

645

$endgroup$

share|cite|improve this question

edited Apr 1 at 17:42

Bernard

124k742117

asked Apr 1 at 16:33

Takeshi GoudaTakeshi Gouda

645

share|cite|improve this question

edited Apr 1 at 17:42

Bernard

124k742117

asked Apr 1 at 16:33

Takeshi GoudaTakeshi Gouda

645

edited Apr 1 at 17:42

Bernard

124k742117

edited Apr 1 at 17:42

Bernard

124k742117

asked Apr 1 at 16:33

Takeshi GoudaTakeshi Gouda

645

asked Apr 1 at 16:33

Takeshi GoudaTakeshi Gouda

645

1 Answer
1

active

oldest

votes

1

$begingroup$

Recall that the commutator map $$(S, T) mapsto [S, T] := S circ T - T circ S$$ defines a Lie algebra structure on the vector space $operatornameEnd(V)$ of linear maps $V to V$. We can identify this Lie algebra with the Lie algebra $mathfrakgl(V)$ of the Lie group $GL(V)$ of invertible linear transformations of $V$. (If we choose a basis of $V$, we can identify both of these with the space $M(n, Bbb R)$ of $n times n$ real matrices, where $n := dim V$, endowed with the commutator $$(A, B) mapsto A B - B A$$ of matrices.)

In particular, we can regard any linear transformation $T in operatornameEnd(V)$ as an element of $mathfrakgl(V)$, and so it determines a left-invariant vector field $tilde T$ on $GL(V)$ characterized by $$tilde T_1 = T .$$

Now, the map $gamma : Bbb R to GL(V)$ defined by $gamma : t mapsto operatornameExp(t T)$ is a smooth curve in $GL(V)$, and unwinding definitions shows that it is an integral curve of $tilde T$, that is, that $$gamma'(t) = tilde T_gamma(t)$$ for all $t$.

share|cite|improve this answer

edited Apr 1 at 17:55

answered Apr 1 at 17:50

TravisTravis

64.5k769151

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer, I think I'm almost getting it... Could you possibly elaborate on how to see that $End(V)$ is identified with $gl(V)$? I know that $gl(V)$ is isomorphic to the tangent space to $GL(V)$ at identity, but how do you identify this with $End(V)$? Thank you
  $endgroup$
  – Takeshi Gouda
  Apr 2 at 11:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  There are a few ways to see this. One is that $GL(V)$ is an open subset of the vector space $operatornameEnd(V)$, so we can identify $mathfrakgl(V) cong T_1 GL(V)$ with $T_1 operatornameEnd(V)$, but for any vector space $W$ (in particular for $W = operatornameEnd(V)$) and any $w in W$ there is a canonical isomorphism $T_w W cong W$. We can make this more concrete by fixing a basis of $V$, which determines isomorphisms $operatornameEnd(V) cong M(n, Bbb R)$ and $GL(V) cong GL(n, Bbb R)$, where $n := dim V$.
  $endgroup$
  – Travis
  Apr 3 at 0:48
  

  
  
  
  

add a comment | 

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1

$begingroup$

Recall that the commutator map $$(S, T) mapsto [S, T] := S circ T - T circ S$$ defines a Lie algebra structure on the vector space $operatornameEnd(V)$ of linear maps $V to V$. We can identify this Lie algebra with the Lie algebra $mathfrakgl(V)$ of the Lie group $GL(V)$ of invertible linear transformations of $V$. (If we choose a basis of $V$, we can identify both of these with the space $M(n, Bbb R)$ of $n times n$ real matrices, where $n := dim V$, endowed with the commutator $$(A, B) mapsto A B - B A$$ of matrices.)

In particular, we can regard any linear transformation $T in operatornameEnd(V)$ as an element of $mathfrakgl(V)$, and so it determines a left-invariant vector field $tilde T$ on $GL(V)$ characterized by $$tilde T_1 = T .$$

Now, the map $gamma : Bbb R to GL(V)$ defined by $gamma : t mapsto operatornameExp(t T)$ is a smooth curve in $GL(V)$, and unwinding definitions shows that it is an integral curve of $tilde T$, that is, that $$gamma'(t) = tilde T_gamma(t)$$ for all $t$.

share|cite|improve this answer

edited Apr 1 at 17:55

answered Apr 1 at 17:50

TravisTravis

64.5k769151

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer, I think I'm almost getting it... Could you possibly elaborate on how to see that $End(V)$ is identified with $gl(V)$? I know that $gl(V)$ is isomorphic to the tangent space to $GL(V)$ at identity, but how do you identify this with $End(V)$? Thank you
  $endgroup$
  – Takeshi Gouda
  Apr 2 at 11:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  There are a few ways to see this. One is that $GL(V)$ is an open subset of the vector space $operatornameEnd(V)$, so we can identify $mathfrakgl(V) cong T_1 GL(V)$ with $T_1 operatornameEnd(V)$, but for any vector space $W$ (in particular for $W = operatornameEnd(V)$) and any $w in W$ there is a canonical isomorphism $T_w W cong W$. We can make this more concrete by fixing a basis of $V$, which determines isomorphisms $operatornameEnd(V) cong M(n, Bbb R)$ and $GL(V) cong GL(n, Bbb R)$, where $n := dim V$.
  $endgroup$
  – Travis
  Apr 3 at 0:48
  

  
  
  
  

add a comment | 

1

$begingroup$

Recall that the commutator map $$(S, T) mapsto [S, T] := S circ T - T circ S$$ defines a Lie algebra structure on the vector space $operatornameEnd(V)$ of linear maps $V to V$. We can identify this Lie algebra with the Lie algebra $mathfrakgl(V)$ of the Lie group $GL(V)$ of invertible linear transformations of $V$. (If we choose a basis of $V$, we can identify both of these with the space $M(n, Bbb R)$ of $n times n$ real matrices, where $n := dim V$, endowed with the commutator $$(A, B) mapsto A B - B A$$ of matrices.)

In particular, we can regard any linear transformation $T in operatornameEnd(V)$ as an element of $mathfrakgl(V)$, and so it determines a left-invariant vector field $tilde T$ on $GL(V)$ characterized by $$tilde T_1 = T .$$

Now, the map $gamma : Bbb R to GL(V)$ defined by $gamma : t mapsto operatornameExp(t T)$ is a smooth curve in $GL(V)$, and unwinding definitions shows that it is an integral curve of $tilde T$, that is, that $$gamma'(t) = tilde T_gamma(t)$$ for all $t$.

share|cite|improve this answer

edited Apr 1 at 17:55

answered Apr 1 at 17:50

TravisTravis

64.5k769151

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer, I think I'm almost getting it... Could you possibly elaborate on how to see that $End(V)$ is identified with $gl(V)$? I know that $gl(V)$ is isomorphic to the tangent space to $GL(V)$ at identity, but how do you identify this with $End(V)$? Thank you
  $endgroup$
  – Takeshi Gouda
  Apr 2 at 11:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  There are a few ways to see this. One is that $GL(V)$ is an open subset of the vector space $operatornameEnd(V)$, so we can identify $mathfrakgl(V) cong T_1 GL(V)$ with $T_1 operatornameEnd(V)$, but for any vector space $W$ (in particular for $W = operatornameEnd(V)$) and any $w in W$ there is a canonical isomorphism $T_w W cong W$. We can make this more concrete by fixing a basis of $V$, which determines isomorphisms $operatornameEnd(V) cong M(n, Bbb R)$ and $GL(V) cong GL(n, Bbb R)$, where $n := dim V$.
  $endgroup$
  – Travis
  Apr 3 at 0:48
  

  
  
  
  

add a comment | 

$begingroup$

Recall that the commutator map $$(S, T) mapsto [S, T] := S circ T - T circ S$$ defines a Lie algebra structure on the vector space $operatornameEnd(V)$ of linear maps $V to V$. We can identify this Lie algebra with the Lie algebra $mathfrakgl(V)$ of the Lie group $GL(V)$ of invertible linear transformations of $V$. (If we choose a basis of $V$, we can identify both of these with the space $M(n, Bbb R)$ of $n times n$ real matrices, where $n := dim V$, endowed with the commutator $$(A, B) mapsto A B - B A$$ of matrices.)

In particular, we can regard any linear transformation $T in operatornameEnd(V)$ as an element of $mathfrakgl(V)$, and so it determines a left-invariant vector field $tilde T$ on $GL(V)$ characterized by $$tilde T_1 = T .$$

Now, the map $gamma : Bbb R to GL(V)$ defined by $gamma : t mapsto operatornameExp(t T)$ is a smooth curve in $GL(V)$, and unwinding definitions shows that it is an integral curve of $tilde T$, that is, that $$gamma'(t) = tilde T_gamma(t)$$ for all $t$.

share|cite|improve this answer

edited Apr 1 at 17:55

answered Apr 1 at 17:50

TravisTravis

64.5k769151

$endgroup$

share|cite|improve this answer

edited Apr 1 at 17:55

answered Apr 1 at 17:50

TravisTravis

64.5k769151

answered Apr 1 at 17:50

TravisTravis

64.5k769151

answered Apr 1 at 17:50

TravisTravis

64.5k769151