Mixing
A difficulty in understanding an example in Vinberg.
0
$begingroup$
The example is given below:
But I have difficulties in understanding the following:
1- why $V_{0}$ is called $(n-1)-$dimensional subspace, I want a concrete example please?
2- Why if the characteristic of the field $K$ is equal to zero, then $V_{1}$ is not subset of $V_{0}$?
3- I do not understand how he calculated $M((12))x - x$, could anyone clarify this for me?
4- Now I am stucked in the solution of this question (the above example is example 5):
linear-algebra representation-theory symmetric-groups invariant-theory
asked Jan 23 at 23:25
hopefullyhopefully
228114
$endgroup$
add a comment |
0
$begingroup$
The example is given below:
But I have difficulties in understanding the following:
1- why $V_{0}$ is called $(n-1)-$dimensional subspace, I want a concrete example please?
2- Why if the characteristic of the field $K$ is equal to zero, then $V_{1}$ is not subset of $V_{0}$?
3- I do not understand how he calculated $M((12))x - x$, could anyone clarify this for me?
4- Now I am stucked in the solution of this question (the above example is example 5):
linear-algebra representation-theory symmetric-groups invariant-theory
asked Jan 23 at 23:25
hopefullyhopefully
228114
$endgroup$
add a comment |
0
0
0
$begingroup$
The example is given below:
But I have difficulties in understanding the following:
1- why $V_{0}$ is called $(n-1)-$dimensional subspace, I want a concrete example please?
2- Why if the characteristic of the field $K$ is equal to zero, then $V_{1}$ is not subset of $V_{0}$?
3- I do not understand how he calculated $M((12))x - x$, could anyone clarify this for me?
4- Now I am stucked in the solution of this question (the above example is example 5):
linear-algebra representation-theory symmetric-groups invariant-theory
asked Jan 23 at 23:25
hopefullyhopefully
228114
$endgroup$
The example is given below:
But I have difficulties in understanding the following:
1- why $V_{0}$ is called $(n-1)-$dimensional subspace, I want a concrete example please?
2- Why if the characteristic of the field $K$ is equal to zero, then $V_{1}$ is not subset of $V_{0}$?
3- I do not understand how he calculated $M((12))x - x$, could anyone clarify this for me?
4- Now I am stucked in the solution of this question (the above example is example 5):
linear-algebra representation-theory symmetric-groups invariant-theory
linear-algebra representation-theory symmetric-groups invariant-theory
asked Jan 23 at 23:25
hopefullyhopefully
228114
asked Jan 23 at 23:25
hopefullyhopefully
228114
asked Jan 23 at 23:25
hopefullyhopefully
228114
asked Jan 23 at 23:25
hopefullyhopefully
228114
asked Jan 23 at 23:25
hopefullyhopefully
228114
228114
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
1
$begingroup$
I can answer to the first three questions:
$V_0$ is defined by a single linear equation. The general result is this: if a subspace is defined by a system of linear equations of rank $r$ (i.e. the matrix with rows the coefficients of the defining linear equations has rank $r$), this subspace has codimension $r$, which means that in a space of dimension $n$, it has dimension $n-r$. This is because you can determine $r$ coordinates in function of the $n-r$ others (cf. the way to solve linear systems of equations).
$V_0$ is defined by the equation $x_1+x_2+ +dots +x_n=0$, whereas all vectors in $V_1$ satisfy $x_i=x_j :forall i,j$, so that for such a vector, $x_1+x_2+dots+x_n=nx_1$, which is $ne 0$, except for the null vector, or if the characteristic of $K$ is a prime factor of $n$.- If $x=x_1e_1+x_2e_2+sum_{i=3}^n x_ie_i$, we have, by linearity,
$$M((1,2))x=x_1e_2+x_2e_1+sum_{i=3}^n x_ie_i,$$
so that
begin{align}
M((1,2))x-x&=x_1e_2+x_2e_1+sum_{i=3}^n x_ie_i-Bigl(x_1e_1+x_2e_2+sum_{i=3}^n x_ie_iBigr),\
&=x_1e_2+x_2e_1-(x_1e_1+x_2e_2)=x_2e_1-x_1e_1+x_1e_2-x_2e_2\
&=dotsm
end{align}
answered Jan 24 at 0:01
BernardBernard
123k741116
$endgroup$
$begingroup$
what does it mean $sum x_{i}$ = 0 , why we lose one dimension by this assumption could you give a concrete example please?
$endgroup$
– hopefully
Jan 24 at 6:41
$begingroup$
Can you give me a numeric example for $V_{0}$ please?
$endgroup$
– hopefully
Jan 31 at 12:02
1
$begingroup$
Yes: e.g. in $V=mathbf R^3$, $V_0$ is the plane defined by the equation $x+y+z=0$. It also can be defined by $3$ non- collinear points (as any plane). We may take for these $(0,1,-1),:$ $(-1,0,1):$ and $(1,-1,0)$. As to your other question, we lose one dimension because, once we know any $n-1$ coordinates of a point in $V_0$, the $n$-th coordinate can be deduced from this relation. Is that clearer?
$endgroup$
– Bernard
Jan 31 at 14:27
add a comment |
1
$begingroup$
- It is $(n-1)$-dimensional because it has a basis with $n-1$ elements:$${e_1-e_2,e_2-e_3,ldots,e_{n-1}-e_n}.$$
- The set $V_1$ is not a subset of $V_0$ because $e_1+e_2+cdots+e_nin V_1setminus V_0$, since $overbrace{1+1+cdots+1}^{ntext{ terms}}neq0$ (because $K$ has characteristic $0$).
- Did you read the definition if $M(sigma)$? By this definition, $Mbigl((1 2)bigr)(e_1)=e_2$, $Mbigl((1 2)bigr)(e_2)=e_1$, and $Mbigl((1 2)bigr)(e_k)=e_k$ if $k>2$. Therefore, if $x=x_1e_1+x_2e_2+cdots+x_ne_n$, then$$Mbigl((1 2)bigr)(x_1e_1+x_2e_2+cdots+x_ne_n)=x_2e_1+x_1e_2+x_3e_3+cdots+x_ne_n.$$
- I suspect that Vinberg stated some result (about group characters, perhaps) to prove this, but I don't have his book at hand right now.
answered Jan 23 at 23:49
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
$begingroup$
and why it has a basis with n-1 elements?
$endgroup$
– hopefully
Jan 24 at 0:09
$begingroup$
Because that set that I mentioned spans $V_0$, it is linearly independent and it has $n-1$ elements.
$endgroup$
– José Carlos Santos
Jan 24 at 0:12
$begingroup$
Why the set you mentioned spans $V_{0}$ .... why it takes specially this form $e_{i} - e_{i +1}$ ...... why it is not expressed in terms of $e_{i}$'s only?
$endgroup$
– hopefully
Jan 31 at 11:59
1
$begingroup$
It spans $V_0$ because if $vin V_0$, then $v=x_1e_1+x_2e_2+cdots+x_nv_n$, with $x_1+x_2+cdots+x_n=0$, and therefore$$v=x_1(x_1-e_2)+(x_1+x_2)(e_2-e_3)+(x_1+x_2+x_3)(e_3-e_4)+cdots+(x_1+x_2+cdots+x_{n-1})(e_{n-1}-e_n).$$
$endgroup$
– José Carlos Santos
Jan 31 at 12:05
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
I can answer to the first three questions:
$V_0$ is defined by a single linear equation. The general result is this: if a subspace is defined by a system of linear equations of rank $r$ (i.e. the matrix with rows the coefficients of the defining linear equations has rank $r$), this subspace has codimension $r$, which means that in a space of dimension $n$, it has dimension $n-r$. This is because you can determine $r$ coordinates in function of the $n-r$ others (cf. the way to solve linear systems of equations).
$V_0$ is defined by the equation $x_1+x_2+ +dots +x_n=0$, whereas all vectors in $V_1$ satisfy $x_i=x_j :forall i,j$, so that for such a vector, $x_1+x_2+dots+x_n=nx_1$, which is $ne 0$, except for the null vector, or if the characteristic of $K$ is a prime factor of $n$.- If $x=x_1e_1+x_2e_2+sum_{i=3}^n x_ie_i$, we have, by linearity,
$$M((1,2))x=x_1e_2+x_2e_1+sum_{i=3}^n x_ie_i,$$
so that
begin{align}
M((1,2))x-x&=x_1e_2+x_2e_1+sum_{i=3}^n x_ie_i-Bigl(x_1e_1+x_2e_2+sum_{i=3}^n x_ie_iBigr),\
&=x_1e_2+x_2e_1-(x_1e_1+x_2e_2)=x_2e_1-x_1e_1+x_1e_2-x_2e_2\
&=dotsm
end{align}
answered Jan 24 at 0:01
BernardBernard
123k741116
$endgroup$
$begingroup$
what does it mean $sum x_{i}$ = 0 , why we lose one dimension by this assumption could you give a concrete example please?
$endgroup$
– hopefully
Jan 24 at 6:41
$begingroup$
Can you give me a numeric example for $V_{0}$ please?
$endgroup$
– hopefully
Jan 31 at 12:02
1
$begingroup$
Yes: e.g. in $V=mathbf R^3$, $V_0$ is the plane defined by the equation $x+y+z=0$. It also can be defined by $3$ non- collinear points (as any plane). We may take for these $(0,1,-1),:$ $(-1,0,1):$ and $(1,-1,0)$. As to your other question, we lose one dimension because, once we know any $n-1$ coordinates of a point in $V_0$, the $n$-th coordinate can be deduced from this relation. Is that clearer?
$endgroup$
– Bernard
Jan 31 at 14:27
add a comment |
1
$begingroup$
I can answer to the first three questions:
$V_0$ is defined by a single linear equation. The general result is this: if a subspace is defined by a system of linear equations of rank $r$ (i.e. the matrix with rows the coefficients of the defining linear equations has rank $r$), this subspace has codimension $r$, which means that in a space of dimension $n$, it has dimension $n-r$. This is because you can determine $r$ coordinates in function of the $n-r$ others (cf. the way to solve linear systems of equations).
$V_0$ is defined by the equation $x_1+x_2+ +dots +x_n=0$, whereas all vectors in $V_1$ satisfy $x_i=x_j :forall i,j$, so that for such a vector, $x_1+x_2+dots+x_n=nx_1$, which is $ne 0$, except for the null vector, or if the characteristic of $K$ is a prime factor of $n$.- If $x=x_1e_1+x_2e_2+sum_{i=3}^n x_ie_i$, we have, by linearity,
$$M((1,2))x=x_1e_2+x_2e_1+sum_{i=3}^n x_ie_i,$$
so that
begin{align}
M((1,2))x-x&=x_1e_2+x_2e_1+sum_{i=3}^n x_ie_i-Bigl(x_1e_1+x_2e_2+sum_{i=3}^n x_ie_iBigr),\
&=x_1e_2+x_2e_1-(x_1e_1+x_2e_2)=x_2e_1-x_1e_1+x_1e_2-x_2e_2\
&=dotsm
end{align}
answered Jan 24 at 0:01
BernardBernard
123k741116
$endgroup$
$begingroup$
what does it mean $sum x_{i}$ = 0 , why we lose one dimension by this assumption could you give a concrete example please?
$endgroup$
– hopefully
Jan 24 at 6:41
$begingroup$
Can you give me a numeric example for $V_{0}$ please?
$endgroup$
– hopefully
Jan 31 at 12:02
1
$begingroup$
Yes: e.g. in $V=mathbf R^3$, $V_0$ is the plane defined by the equation $x+y+z=0$. It also can be defined by $3$ non- collinear points (as any plane). We may take for these $(0,1,-1),:$ $(-1,0,1):$ and $(1,-1,0)$. As to your other question, we lose one dimension because, once we know any $n-1$ coordinates of a point in $V_0$, the $n$-th coordinate can be deduced from this relation. Is that clearer?
$endgroup$
– Bernard
Jan 31 at 14:27
add a comment |
1
1
1
$begingroup$
I can answer to the first three questions:
$V_0$ is defined by a single linear equation. The general result is this: if a subspace is defined by a system of linear equations of rank $r$ (i.e. the matrix with rows the coefficients of the defining linear equations has rank $r$), this subspace has codimension $r$, which means that in a space of dimension $n$, it has dimension $n-r$. This is because you can determine $r$ coordinates in function of the $n-r$ others (cf. the way to solve linear systems of equations).
$V_0$ is defined by the equation $x_1+x_2+ +dots +x_n=0$, whereas all vectors in $V_1$ satisfy $x_i=x_j :forall i,j$, so that for such a vector, $x_1+x_2+dots+x_n=nx_1$, which is $ne 0$, except for the null vector, or if the characteristic of $K$ is a prime factor of $n$.- If $x=x_1e_1+x_2e_2+sum_{i=3}^n x_ie_i$, we have, by linearity,
$$M((1,2))x=x_1e_2+x_2e_1+sum_{i=3}^n x_ie_i,$$
so that
begin{align}
M((1,2))x-x&=x_1e_2+x_2e_1+sum_{i=3}^n x_ie_i-Bigl(x_1e_1+x_2e_2+sum_{i=3}^n x_ie_iBigr),\
&=x_1e_2+x_2e_1-(x_1e_1+x_2e_2)=x_2e_1-x_1e_1+x_1e_2-x_2e_2\
&=dotsm
end{align}
answered Jan 24 at 0:01
BernardBernard
123k741116
$endgroup$
I can answer to the first three questions:
$V_0$ is defined by a single linear equation. The general result is this: if a subspace is defined by a system of linear equations of rank $r$ (i.e. the matrix with rows the coefficients of the defining linear equations has rank $r$), this subspace has codimension $r$, which means that in a space of dimension $n$, it has dimension $n-r$. This is because you can determine $r$ coordinates in function of the $n-r$ others (cf. the way to solve linear systems of equations).
$V_0$ is defined by the equation $x_1+x_2+ +dots +x_n=0$, whereas all vectors in $V_1$ satisfy $x_i=x_j :forall i,j$, so that for such a vector, $x_1+x_2+dots+x_n=nx_1$, which is $ne 0$, except for the null vector, or if the characteristic of $K$ is a prime factor of $n$.- If $x=x_1e_1+x_2e_2+sum_{i=3}^n x_ie_i$, we have, by linearity,
$$M((1,2))x=x_1e_2+x_2e_1+sum_{i=3}^n x_ie_i,$$
so that
begin{align}
M((1,2))x-x&=x_1e_2+x_2e_1+sum_{i=3}^n x_ie_i-Bigl(x_1e_1+x_2e_2+sum_{i=3}^n x_ie_iBigr),\
&=x_1e_2+x_2e_1-(x_1e_1+x_2e_2)=x_2e_1-x_1e_1+x_1e_2-x_2e_2\
&=dotsm
end{align}
answered Jan 24 at 0:01
BernardBernard
123k741116
edited Jan 24 at 0:13
answered Jan 24 at 0:01
BernardBernard
123k741116
answered Jan 24 at 0:01
BernardBernard
123k741116
answered Jan 24 at 0:01
BernardBernard
123k741116
123k741116
$begingroup$
what does it mean $sum x_{i}$ = 0 , why we lose one dimension by this assumption could you give a concrete example please?
$endgroup$
– hopefully
Jan 24 at 6:41
$begingroup$
Can you give me a numeric example for $V_{0}$ please?
$endgroup$
– hopefully
Jan 31 at 12:02
1
$begingroup$
Yes: e.g. in $V=mathbf R^3$, $V_0$ is the plane defined by the equation $x+y+z=0$. It also can be defined by $3$ non- collinear points (as any plane). We may take for these $(0,1,-1),:$ $(-1,0,1):$ and $(1,-1,0)$. As to your other question, we lose one dimension because, once we know any $n-1$ coordinates of a point in $V_0$, the $n$-th coordinate can be deduced from this relation. Is that clearer?
$endgroup$
– Bernard
Jan 31 at 14:27
add a comment |
$begingroup$
what does it mean $sum x_{i}$ = 0 , why we lose one dimension by this assumption could you give a concrete example please?
$endgroup$
– hopefully
Jan 24 at 6:41
$begingroup$
Can you give me a numeric example for $V_{0}$ please?
$endgroup$
– hopefully
Jan 31 at 12:02
1
$begingroup$
Yes: e.g. in $V=mathbf R^3$, $V_0$ is the plane defined by the equation $x+y+z=0$. It also can be defined by $3$ non- collinear points (as any plane). We may take for these $(0,1,-1),:$ $(-1,0,1):$ and $(1,-1,0)$. As to your other question, we lose one dimension because, once we know any $n-1$ coordinates of a point in $V_0$, the $n$-th coordinate can be deduced from this relation. Is that clearer?
$endgroup$
– Bernard
Jan 31 at 14:27
$begingroup$
what does it mean $sum x_{i}$ = 0 , why we lose one dimension by this assumption could you give a concrete example please?
$endgroup$
– hopefully
Jan 24 at 6:41
$begingroup$
what does it mean $sum x_{i}$ = 0 , why we lose one dimension by this assumption could you give a concrete example please?
$endgroup$
– hopefully
Jan 24 at 6:41
$begingroup$
Can you give me a numeric example for $V_{0}$ please?
$endgroup$
– hopefully
Jan 31 at 12:02
$begingroup$
Can you give me a numeric example for $V_{0}$ please?
$endgroup$
– hopefully
Jan 31 at 12:02
1
1
$begingroup$
Yes: e.g. in $V=mathbf R^3$, $V_0$ is the plane defined by the equation $x+y+z=0$. It also can be defined by $3$ non- collinear points (as any plane). We may take for these $(0,1,-1),:$ $(-1,0,1):$ and $(1,-1,0)$. As to your other question, we lose one dimension because, once we know any $n-1$ coordinates of a point in $V_0$, the $n$-th coordinate can be deduced from this relation. Is that clearer?
$endgroup$
– Bernard
Jan 31 at 14:27
$begingroup$
Yes: e.g. in $V=mathbf R^3$, $V_0$ is the plane defined by the equation $x+y+z=0$. It also can be defined by $3$ non- collinear points (as any plane). We may take for these $(0,1,-1),:$ $(-1,0,1):$ and $(1,-1,0)$. As to your other question, we lose one dimension because, once we know any $n-1$ coordinates of a point in $V_0$, the $n$-th coordinate can be deduced from this relation. Is that clearer?
$endgroup$
– Bernard
Jan 31 at 14:27
add a comment |
1
$begingroup$
- It is $(n-1)$-dimensional because it has a basis with $n-1$ elements:$${e_1-e_2,e_2-e_3,ldots,e_{n-1}-e_n}.$$
- The set $V_1$ is not a subset of $V_0$ because $e_1+e_2+cdots+e_nin V_1setminus V_0$, since $overbrace{1+1+cdots+1}^{ntext{ terms}}neq0$ (because $K$ has characteristic $0$).
- Did you read the definition if $M(sigma)$? By this definition, $Mbigl((1 2)bigr)(e_1)=e_2$, $Mbigl((1 2)bigr)(e_2)=e_1$, and $Mbigl((1 2)bigr)(e_k)=e_k$ if $k>2$. Therefore, if $x=x_1e_1+x_2e_2+cdots+x_ne_n$, then$$Mbigl((1 2)bigr)(x_1e_1+x_2e_2+cdots+x_ne_n)=x_2e_1+x_1e_2+x_3e_3+cdots+x_ne_n.$$
- I suspect that Vinberg stated some result (about group characters, perhaps) to prove this, but I don't have his book at hand right now.
answered Jan 23 at 23:49
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
$begingroup$
and why it has a basis with n-1 elements?
$endgroup$
– hopefully
Jan 24 at 0:09
$begingroup$
Because that set that I mentioned spans $V_0$, it is linearly independent and it has $n-1$ elements.
$endgroup$
– José Carlos Santos
Jan 24 at 0:12
$begingroup$
Why the set you mentioned spans $V_{0}$ .... why it takes specially this form $e_{i} - e_{i +1}$ ...... why it is not expressed in terms of $e_{i}$'s only?
$endgroup$
– hopefully
Jan 31 at 11:59
1
$begingroup$
It spans $V_0$ because if $vin V_0$, then $v=x_1e_1+x_2e_2+cdots+x_nv_n$, with $x_1+x_2+cdots+x_n=0$, and therefore$$v=x_1(x_1-e_2)+(x_1+x_2)(e_2-e_3)+(x_1+x_2+x_3)(e_3-e_4)+cdots+(x_1+x_2+cdots+x_{n-1})(e_{n-1}-e_n).$$
$endgroup$
– José Carlos Santos
Jan 31 at 12:05
add a comment |
1
$begingroup$
- It is $(n-1)$-dimensional because it has a basis with $n-1$ elements:$${e_1-e_2,e_2-e_3,ldots,e_{n-1}-e_n}.$$
- The set $V_1$ is not a subset of $V_0$ because $e_1+e_2+cdots+e_nin V_1setminus V_0$, since $overbrace{1+1+cdots+1}^{ntext{ terms}}neq0$ (because $K$ has characteristic $0$).
- Did you read the definition if $M(sigma)$? By this definition, $Mbigl((1 2)bigr)(e_1)=e_2$, $Mbigl((1 2)bigr)(e_2)=e_1$, and $Mbigl((1 2)bigr)(e_k)=e_k$ if $k>2$. Therefore, if $x=x_1e_1+x_2e_2+cdots+x_ne_n$, then$$Mbigl((1 2)bigr)(x_1e_1+x_2e_2+cdots+x_ne_n)=x_2e_1+x_1e_2+x_3e_3+cdots+x_ne_n.$$
- I suspect that Vinberg stated some result (about group characters, perhaps) to prove this, but I don't have his book at hand right now.
answered Jan 23 at 23:49
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
$begingroup$
and why it has a basis with n-1 elements?
$endgroup$
– hopefully
Jan 24 at 0:09
$begingroup$
Because that set that I mentioned spans $V_0$, it is linearly independent and it has $n-1$ elements.
$endgroup$
– José Carlos Santos
Jan 24 at 0:12
$begingroup$
Why the set you mentioned spans $V_{0}$ .... why it takes specially this form $e_{i} - e_{i +1}$ ...... why it is not expressed in terms of $e_{i}$'s only?
$endgroup$
– hopefully
Jan 31 at 11:59
1
$begingroup$
It spans $V_0$ because if $vin V_0$, then $v=x_1e_1+x_2e_2+cdots+x_nv_n$, with $x_1+x_2+cdots+x_n=0$, and therefore$$v=x_1(x_1-e_2)+(x_1+x_2)(e_2-e_3)+(x_1+x_2+x_3)(e_3-e_4)+cdots+(x_1+x_2+cdots+x_{n-1})(e_{n-1}-e_n).$$
$endgroup$
– José Carlos Santos
Jan 31 at 12:05
add a comment |
1
1
1
$begingroup$
- It is $(n-1)$-dimensional because it has a basis with $n-1$ elements:$${e_1-e_2,e_2-e_3,ldots,e_{n-1}-e_n}.$$
- The set $V_1$ is not a subset of $V_0$ because $e_1+e_2+cdots+e_nin V_1setminus V_0$, since $overbrace{1+1+cdots+1}^{ntext{ terms}}neq0$ (because $K$ has characteristic $0$).
- Did you read the definition if $M(sigma)$? By this definition, $Mbigl((1 2)bigr)(e_1)=e_2$, $Mbigl((1 2)bigr)(e_2)=e_1$, and $Mbigl((1 2)bigr)(e_k)=e_k$ if $k>2$. Therefore, if $x=x_1e_1+x_2e_2+cdots+x_ne_n$, then$$Mbigl((1 2)bigr)(x_1e_1+x_2e_2+cdots+x_ne_n)=x_2e_1+x_1e_2+x_3e_3+cdots+x_ne_n.$$
- I suspect that Vinberg stated some result (about group characters, perhaps) to prove this, but I don't have his book at hand right now.
answered Jan 23 at 23:49
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
- It is $(n-1)$-dimensional because it has a basis with $n-1$ elements:$${e_1-e_2,e_2-e_3,ldots,e_{n-1}-e_n}.$$
- The set $V_1$ is not a subset of $V_0$ because $e_1+e_2+cdots+e_nin V_1setminus V_0$, since $overbrace{1+1+cdots+1}^{ntext{ terms}}neq0$ (because $K$ has characteristic $0$).
- Did you read the definition if $M(sigma)$? By this definition, $Mbigl((1 2)bigr)(e_1)=e_2$, $Mbigl((1 2)bigr)(e_2)=e_1$, and $Mbigl((1 2)bigr)(e_k)=e_k$ if $k>2$. Therefore, if $x=x_1e_1+x_2e_2+cdots+x_ne_n$, then$$Mbigl((1 2)bigr)(x_1e_1+x_2e_2+cdots+x_ne_n)=x_2e_1+x_1e_2+x_3e_3+cdots+x_ne_n.$$
- I suspect that Vinberg stated some result (about group characters, perhaps) to prove this, but I don't have his book at hand right now.
answered Jan 23 at 23:49
José Carlos SantosJosé Carlos Santos
167k22132235
answered Jan 23 at 23:49
José Carlos SantosJosé Carlos Santos
167k22132235
answered Jan 23 at 23:49
José Carlos SantosJosé Carlos Santos
167k22132235
answered Jan 23 at 23:49
José Carlos SantosJosé Carlos Santos
167k22132235
167k22132235
$begingroup$
and why it has a basis with n-1 elements?
$endgroup$
– hopefully
Jan 24 at 0:09
$begingroup$
Because that set that I mentioned spans $V_0$, it is linearly independent and it has $n-1$ elements.
$endgroup$
– José Carlos Santos
Jan 24 at 0:12
$begingroup$
Why the set you mentioned spans $V_{0}$ .... why it takes specially this form $e_{i} - e_{i +1}$ ...... why it is not expressed in terms of $e_{i}$'s only?
$endgroup$
– hopefully
Jan 31 at 11:59
1
$begingroup$
It spans $V_0$ because if $vin V_0$, then $v=x_1e_1+x_2e_2+cdots+x_nv_n$, with $x_1+x_2+cdots+x_n=0$, and therefore$$v=x_1(x_1-e_2)+(x_1+x_2)(e_2-e_3)+(x_1+x_2+x_3)(e_3-e_4)+cdots+(x_1+x_2+cdots+x_{n-1})(e_{n-1}-e_n).$$
$endgroup$
– José Carlos Santos
Jan 31 at 12:05
add a comment |
$begingroup$
and why it has a basis with n-1 elements?
$endgroup$
– hopefully
Jan 24 at 0:09
$begingroup$
Because that set that I mentioned spans $V_0$, it is linearly independent and it has $n-1$ elements.
$endgroup$
– José Carlos Santos
Jan 24 at 0:12
$begingroup$
Why the set you mentioned spans $V_{0}$ .... why it takes specially this form $e_{i} - e_{i +1}$ ...... why it is not expressed in terms of $e_{i}$'s only?
$endgroup$
– hopefully
Jan 31 at 11:59
1
$begingroup$
It spans $V_0$ because if $vin V_0$, then $v=x_1e_1+x_2e_2+cdots+x_nv_n$, with $x_1+x_2+cdots+x_n=0$, and therefore$$v=x_1(x_1-e_2)+(x_1+x_2)(e_2-e_3)+(x_1+x_2+x_3)(e_3-e_4)+cdots+(x_1+x_2+cdots+x_{n-1})(e_{n-1}-e_n).$$
$endgroup$
– José Carlos Santos
Jan 31 at 12:05
$begingroup$
and why it has a basis with n-1 elements?
$endgroup$
– hopefully
Jan 24 at 0:09
$begingroup$
and why it has a basis with n-1 elements?
$endgroup$
– hopefully
Jan 24 at 0:09
$begingroup$
Because that set that I mentioned spans $V_0$, it is linearly independent and it has $n-1$ elements.
$endgroup$
– José Carlos Santos
Jan 24 at 0:12
$begingroup$
Because that set that I mentioned spans $V_0$, it is linearly independent and it has $n-1$ elements.
$endgroup$
– José Carlos Santos
Jan 24 at 0:12
$begingroup$
Why the set you mentioned spans $V_{0}$ .... why it takes specially this form $e_{i} - e_{i +1}$ ...... why it is not expressed in terms of $e_{i}$'s only?
$endgroup$
– hopefully
Jan 31 at 11:59
$begingroup$
Why the set you mentioned spans $V_{0}$ .... why it takes specially this form $e_{i} - e_{i +1}$ ...... why it is not expressed in terms of $e_{i}$'s only?
$endgroup$
– hopefully
Jan 31 at 11:59
1
1
$begingroup$
It spans $V_0$ because if $vin V_0$, then $v=x_1e_1+x_2e_2+cdots+x_nv_n$, with $x_1+x_2+cdots+x_n=0$, and therefore$$v=x_1(x_1-e_2)+(x_1+x_2)(e_2-e_3)+(x_1+x_2+x_3)(e_3-e_4)+cdots+(x_1+x_2+cdots+x_{n-1})(e_{n-1}-e_n).$$
$endgroup$
– José Carlos Santos
Jan 31 at 12:05
$begingroup$
It spans $V_0$ because if $vin V_0$, then $v=x_1e_1+x_2e_2+cdots+x_nv_n$, with $x_1+x_2+cdots+x_n=0$, and therefore$$v=x_1(x_1-e_2)+(x_1+x_2)(e_2-e_3)+(x_1+x_2+x_3)(e_3-e_4)+cdots+(x_1+x_2+cdots+x_{n-1})(e_{n-1}-e_n).$$
$endgroup$
– José Carlos Santos
Jan 31 at 12:05
add a comment |
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