[SOLVED] Pseudo orthogonal group

Marc Nardmann

Arnold Neumaier wrote:

> In particular, SO(p,q) is connected if the product pq is even,
> and has two connected components otherwise.


This is not true. SO(p,q) has always two connected components if pq>0.
At least with the standard definition of SO(p,q) that I repeat in
another message on this thread. In fact, it is very easy to see that
there are always *at least* two connected components; cf. my other message.

> The book by
> R. Gilmore,
> Lie groups, Lie algebras, and some of their applications
> Wiley, New York 1974
> contains a lot of material about specific classical groups.
> The above result is stated there (without proof) on p. 199.


I haven't looked it up. Maybe a clash of notation?


-- Marc Nardmann

Marc Nardmann

Arnold Neumaier wrote:

> In particular, SO(p,q) is connected if the product pq is even,
> and has two connected components otherwise.


This is not true. SO(p,q) has always two connected components if pq>0.
At least with the standard definition of SO(p,q) that I repeat in
another message on this thread. In fact, it is very easy to see that
there are always *at least* two connected components; cf. my other message.

> The book by
> R. Gilmore,
> Lie groups, Lie algebras, and some of their applications
> Wiley, New York 1974
> contains a lot of material about specific classical groups.
> The above result is stated there (without proof) on p. 199.


I haven't looked it up. Maybe a clash of notation?


-- Marc Nardmann

Marc Nardmann

Arnold Neumaier wrote:

> In particular, SO(p,q) is connected if the product pq is even,
> and has two connected components otherwise.


This is not true. SO(p,q) has always two connected components if pq>0.
At least with the standard definition of SO(p,q) that I repeat in
another message on this thread. In fact, it is very easy to see that
there are always *at least* two connected components; cf. my other message.

> The book by
> R. Gilmore,
> Lie groups, Lie algebras, and some of their applications
> Wiley, New York 1974
> contains a lot of material about specific classical groups.
> The above result is stated there (without proof) on p. 199.


I haven't looked it up. Maybe a clash of notation?


-- Marc Nardmann

Marc Nardmann

Arnold Neumaier wrote:

> In particular, SO(p,q) is connected if the product pq is even,
> and has two connected components otherwise.


This is not true. SO(p,q) has always two connected components if pq>0.
At least with the standard definition of SO(p,q) that I repeat in
another message on this thread. In fact, it is very easy to see that
there are always *at least* two connected components; cf. my other message.

> The book by
> R. Gilmore,
> Lie groups, Lie algebras, and some of their applications
> Wiley, New York 1974
> contains a lot of material about specific classical groups.
> The above result is stated there (without proof) on p. 199.


I haven't looked it up. Maybe a clash of notation?


-- Marc Nardmann

Marc Nardmann

Arnold Neumaier wrote:

> In particular, SO(p,q) is connected if the product pq is even,
> and has two connected components otherwise.


This is not true. SO(p,q) has always two connected components if pq>0.
At least with the standard definition of SO(p,q) that I repeat in
another message on this thread. In fact, it is very easy to see that
there are always *at least* two connected components; cf. my other message.

> The book by
> R. Gilmore,
> Lie groups, Lie algebras, and some of their applications
> Wiley, New York 1974
> contains a lot of material about specific classical groups.
> The above result is stated there (without proof) on p. 199.


I haven't looked it up. Maybe a clash of notation?


-- Marc Nardmann

Marc Nardmann

Arnold Neumaier wrote:

> In particular, SO(p,q) is connected if the product pq is even,
> and has two connected components otherwise.


This is not true. SO(p,q) has always two connected components if pq>0.
At least with the standard definition of SO(p,q) that I repeat in
another message on this thread. In fact, it is very easy to see that
there are always *at least* two connected components; cf. my other message.

> The book by
> R. Gilmore,
> Lie groups, Lie algebras, and some of their applications
> Wiley, New York 1974
> contains a lot of material about specific classical groups.
> The above result is stated there (without proof) on p. 199.


I haven't looked it up. Maybe a clash of notation?


-- Marc Nardmann

Marc Nardmann

Arnold Neumaier wrote:

> In particular, SO(p,q) is connected if the product pq is even,
> and has two connected components otherwise.


This is not true. SO(p,q) has always two connected components if pq>0.
At least with the standard definition of SO(p,q) that I repeat in
another message on this thread. In fact, it is very easy to see that
there are always *at least* two connected components; cf. my other message.

> The book by
> R. Gilmore,
> Lie groups, Lie algebras, and some of their applications
> Wiley, New York 1974
> contains a lot of material about specific classical groups.
> The above result is stated there (without proof) on p. 199.


I haven't looked it up. Maybe a clash of notation?


-- Marc Nardmann

Marc Nardmann

Arnold Neumaier wrote:

> In particular, SO(p,q) is connected if the product pq is even,
> and has two connected components otherwise.


This is not true. SO(p,q) has always two connected components if pq>0.
At least with the standard definition of SO(p,q) that I repeat in
another message on this thread. In fact, it is very easy to see that
there are always *at least* two connected components; cf. my other message.

> The book by
> R. Gilmore,
> Lie groups, Lie algebras, and some of their applications
> Wiley, New York 1974
> contains a lot of material about specific classical groups.
> The above result is stated there (without proof) on p. 199.


I haven't looked it up. Maybe a clash of notation?


-- Marc Nardmann

Marc Nardmann

Arnold Neumaier wrote:

> In particular, SO(p,q) is connected if the product pq is even,
> and has two connected components otherwise.


This is not true. SO(p,q) has always two connected components if pq>0.
At least with the standard definition of SO(p,q) that I repeat in
another message on this thread. In fact, it is very easy to see that
there are always *at least* two connected components; cf. my other message.

> The book by
> R. Gilmore,
> Lie groups, Lie algebras, and some of their applications
> Wiley, New York 1974
> contains a lot of material about specific classical groups.
> The above result is stated there (without proof) on p. 199.


I haven't looked it up. Maybe a clash of notation?


-- Marc Nardmann

Marc Nardmann

Arnold Neumaier wrote:

> In particular, SO(p,q) is connected if the product pq is even,
> and has two connected components otherwise.


This is not true. SO(p,q) has always two connected components if pq>0.
At least with the standard definition of SO(p,q) that I repeat in
another message on this thread. In fact, it is very easy to see that
there are always *at least* two connected components; cf. my other message.

> The book by
> R. Gilmore,
> Lie groups, Lie algebras, and some of their applications
> Wiley, New York 1974
> contains a lot of material about specific classical groups.
> The above result is stated there (without proof) on p. 199.


I haven't looked it up. Maybe a clash of notation?


-- Marc Nardmann

Marc Nardmann

Arnold Neumaier wrote:

> In particular, SO(p,q) is connected if the product pq is even,
> and has two connected components otherwise.


This is not true. SO(p,q) has always two connected components if pq>0.
At least with the standard definition of SO(p,q) that I repeat in
another message on this thread. In fact, it is very easy to see that
there are always *at least* two connected components; cf. my other message.

> The book by
> R. Gilmore,
> Lie groups, Lie algebras, and some of their applications
> Wiley, New York 1974
> contains a lot of material about specific classical groups.
> The above result is stated there (without proof) on p. 199.


I haven't looked it up. Maybe a clash of notation?


-- Marc Nardmann

Marc Nardmann

I wrote:

>Now we show that for every p x q matrix F (e.g. F=B(t)), the matrix 1-
>F(1+F'F)^{-1}F' is positive definite: Let v be in R^p. Then v = Fx +w
>for some x in R^q and some w in R^p which is orthogonal to the image of
>F. Let G denote the positive semidefinite matrix F'F. We have to show that
>
><v,v> - <F(1+G)^{-1}F'v,v>
>
>is positive if v is nonzero. Since F'w = 0, we obtain
>
><v,v> - <F(1+G)^{-1}F'v,v> = <w,w> + <Fx,Fx> - <F(1+G)^{-1}F'Fx, Fx> ,
>
>so it suffices to prove that <Gx,x> - <(1+G)^{-1}Gx, Gx> is positive if
>x is nonzero. Let x be nonzero, y = (1+G)^{-1}x. Then
>
>0 < <Gy, y> + <GGy, y>
> = <G(1+G)y, y>
> = <Gx, (1+G)^{-1}x>
> = <Gx, (1+G)^{-1}(1+G)x -(1+G)^{-1}Gx>
> = <Gx, x> - <Gx, (1+G)^{-1}Gx> ,
>
>as claimed. Thus 1- F(1+F'F)^{-1}F' is indeed positive definite.
>
>

Oops, not quite. What I should have said is this:

> [...] so it suffices to prove that <Gx,x> - <(1+G)^{-1}Gx, Gx> is
> positive if
> Fx is nonzero. Let Fx be nonzero, y = (1+G)^{-1}x. Then Gx is nonzero
> because <Gx,x> = <Fx,Fx> is nonzero. Hence
>
> <Gy,Gx> = <G(1+G)^{-1}x, Gx> = <(1+G)^{-1}Gx,Gx> > 0 ,
>
> so Gy is nonzero. Thus
>
> 0 < <Gy,y> + <Gy,Gy> = <Gy, y> + <GGy, y>
>
> [...]


Near the end of the message, there was a typo:

>C(1) sqrt[1+C(1)'C(1)]^{-1} = D'^{-1}B' U = C sqrt[1+C'C] , hence
>

should be:

>C(1) sqrt[1+C(1)'C(1)]^{-1} = D'^{-1}B' U = C sqrt[1+C'C]^{-1} , hence
>


-- Marc Nardmann

Marc Nardmann

I wrote:

>Now we show that for every p x q matrix F (e.g. F=B(t)), the matrix 1-
>F(1+F'F)^{-1}F' is positive definite: Let v be in R^p. Then v = Fx +w
>for some x in R^q and some w in R^p which is orthogonal to the image of
>F. Let G denote the positive semidefinite matrix F'F. We have to show that
>
><v,v> - <F(1+G)^{-1}F'v,v>
>
>is positive if v is nonzero. Since F'w = 0, we obtain
>
><v,v> - <F(1+G)^{-1}F'v,v> = <w,w> + <Fx,Fx> - <F(1+G)^{-1}F'Fx, Fx> ,
>
>so it suffices to prove that <Gx,x> - <(1+G)^{-1}Gx, Gx> is positive if
>x is nonzero. Let x be nonzero, y = (1+G)^{-1}x. Then
>
>0 < <Gy, y> + <GGy, y>
> = <G(1+G)y, y>
> = <Gx, (1+G)^{-1}x>
> = <Gx, (1+G)^{-1}(1+G)x -(1+G)^{-1}Gx>
> = <Gx, x> - <Gx, (1+G)^{-1}Gx> ,
>
>as claimed. Thus 1- F(1+F'F)^{-1}F' is indeed positive definite.
>
>

Oops, not quite. What I should have said is this:

> [...] so it suffices to prove that <Gx,x> - <(1+G)^{-1}Gx, Gx> is
> positive if
> Fx is nonzero. Let Fx be nonzero, y = (1+G)^{-1}x. Then Gx is nonzero
> because <Gx,x> = <Fx,Fx> is nonzero. Hence
>
> <Gy,Gx> = <G(1+G)^{-1}x, Gx> = <(1+G)^{-1}Gx,Gx> > 0 ,
>
> so Gy is nonzero. Thus
>
> 0 < <Gy,y> + <Gy,Gy> = <Gy, y> + <GGy, y>
>
> [...]


Near the end of the message, there was a typo:

>C(1) sqrt[1+C(1)'C(1)]^{-1} = D'^{-1}B' U = C sqrt[1+C'C] , hence
>

should be:

>C(1) sqrt[1+C(1)'C(1)]^{-1} = D'^{-1}B' U = C sqrt[1+C'C]^{-1} , hence
>


-- Marc Nardmann

Marc Nardmann

I wrote:

>Now we show that for every p x q matrix F (e.g. F=B(t)), the matrix 1-
>F(1+F'F)^{-1}F' is positive definite: Let v be in R^p. Then v = Fx +w
>for some x in R^q and some w in R^p which is orthogonal to the image of
>F. Let G denote the positive semidefinite matrix F'F. We have to show that
>
><v,v> - <F(1+G)^{-1}F'v,v>
>
>is positive if v is nonzero. Since F'w = 0, we obtain
>
><v,v> - <F(1+G)^{-1}F'v,v> = <w,w> + <Fx,Fx> - <F(1+G)^{-1}F'Fx, Fx> ,
>
>so it suffices to prove that <Gx,x> - <(1+G)^{-1}Gx, Gx> is positive if
>x is nonzero. Let x be nonzero, y = (1+G)^{-1}x. Then
>
>0 < <Gy, y> + <GGy, y>
> = <G(1+G)y, y>
> = <Gx, (1+G)^{-1}x>
> = <Gx, (1+G)^{-1}(1+G)x -(1+G)^{-1}Gx>
> = <Gx, x> - <Gx, (1+G)^{-1}Gx> ,
>
>as claimed. Thus 1- F(1+F'F)^{-1}F' is indeed positive definite.
>
>

Oops, not quite. What I should have said is this:

> [...] so it suffices to prove that <Gx,x> - <(1+G)^{-1}Gx, Gx> is
> positive if
> Fx is nonzero. Let Fx be nonzero, y = (1+G)^{-1}x. Then Gx is nonzero
> because <Gx,x> = <Fx,Fx> is nonzero. Hence
>
> <Gy,Gx> = <G(1+G)^{-1}x, Gx> = <(1+G)^{-1}Gx,Gx> > 0 ,
>
> so Gy is nonzero. Thus
>
> 0 < <Gy,y> + <Gy,Gy> = <Gy, y> + <GGy, y>
>
> [...]


Near the end of the message, there was a typo:

>C(1) sqrt[1+C(1)'C(1)]^{-1} = D'^{-1}B' U = C sqrt[1+C'C] , hence
>

should be:

>C(1) sqrt[1+C(1)'C(1)]^{-1} = D'^{-1}B' U = C sqrt[1+C'C]^{-1} , hence
>


-- Marc Nardmann

Marc Nardmann

I wrote:

>Now we show that for every p x q matrix F (e.g. F=B(t)), the matrix 1-
>F(1+F'F)^{-1}F' is positive definite: Let v be in R^p. Then v = Fx +w
>for some x in R^q and some w in R^p which is orthogonal to the image of
>F. Let G denote the positive semidefinite matrix F'F. We have to show that
>
><v,v> - <F(1+G)^{-1}F'v,v>
>
>is positive if v is nonzero. Since F'w = 0, we obtain
>
><v,v> - <F(1+G)^{-1}F'v,v> = <w,w> + <Fx,Fx> - <F(1+G)^{-1}F'Fx, Fx> ,
>
>so it suffices to prove that <Gx,x> - <(1+G)^{-1}Gx, Gx> is positive if
>x is nonzero. Let x be nonzero, y = (1+G)^{-1}x. Then
>
>0 < <Gy, y> + <GGy, y>
> = <G(1+G)y, y>
> = <Gx, (1+G)^{-1}x>
> = <Gx, (1+G)^{-1}(1+G)x -(1+G)^{-1}Gx>
> = <Gx, x> - <Gx, (1+G)^{-1}Gx> ,
>
>as claimed. Thus 1- F(1+F'F)^{-1}F' is indeed positive definite.
>
>

Oops, not quite. What I should have said is this:

> [...] so it suffices to prove that <Gx,x> - <(1+G)^{-1}Gx, Gx> is
> positive if
> Fx is nonzero. Let Fx be nonzero, y = (1+G)^{-1}x. Then Gx is nonzero
> because <Gx,x> = <Fx,Fx> is nonzero. Hence
>
> <Gy,Gx> = <G(1+G)^{-1}x, Gx> = <(1+G)^{-1}Gx,Gx> > 0 ,
>
> so Gy is nonzero. Thus
>
> 0 < <Gy,y> + <Gy,Gy> = <Gy, y> + <GGy, y>
>
> [...]


Near the end of the message, there was a typo:

>C(1) sqrt[1+C(1)'C(1)]^{-1} = D'^{-1}B' U = C sqrt[1+C'C] , hence
>

should be:

>C(1) sqrt[1+C(1)'C(1)]^{-1} = D'^{-1}B' U = C sqrt[1+C'C]^{-1} , hence
>


-- Marc Nardmann

Marc Nardmann

I wrote:

>Now we show that for every p x q matrix F (e.g. F=B(t)), the matrix 1-
>F(1+F'F)^{-1}F' is positive definite: Let v be in R^p. Then v = Fx +w
>for some x in R^q and some w in R^p which is orthogonal to the image of
>F. Let G denote the positive semidefinite matrix F'F. We have to show that
>
><v,v> - <F(1+G)^{-1}F'v,v>
>
>is positive if v is nonzero. Since F'w = 0, we obtain
>
><v,v> - <F(1+G)^{-1}F'v,v> = <w,w> + <Fx,Fx> - <F(1+G)^{-1}F'Fx, Fx> ,
>
>so it suffices to prove that <Gx,x> - <(1+G)^{-1}Gx, Gx> is positive if
>x is nonzero. Let x be nonzero, y = (1+G)^{-1}x. Then
>
>0 < <Gy, y> + <GGy, y>
> = <G(1+G)y, y>
> = <Gx, (1+G)^{-1}x>
> = <Gx, (1+G)^{-1}(1+G)x -(1+G)^{-1}Gx>
> = <Gx, x> - <Gx, (1+G)^{-1}Gx> ,
>
>as claimed. Thus 1- F(1+F'F)^{-1}F' is indeed positive definite.
>
>

Oops, not quite. What I should have said is this:

> [...] so it suffices to prove that <Gx,x> - <(1+G)^{-1}Gx, Gx> is
> positive if
> Fx is nonzero. Let Fx be nonzero, y = (1+G)^{-1}x. Then Gx is nonzero
> because <Gx,x> = <Fx,Fx> is nonzero. Hence
>
> <Gy,Gx> = <G(1+G)^{-1}x, Gx> = <(1+G)^{-1}Gx,Gx> > 0 ,
>
> so Gy is nonzero. Thus
>
> 0 < <Gy,y> + <Gy,Gy> = <Gy, y> + <GGy, y>
>
> [...]


Near the end of the message, there was a typo:

>C(1) sqrt[1+C(1)'C(1)]^{-1} = D'^{-1}B' U = C sqrt[1+C'C] , hence
>

should be:

>C(1) sqrt[1+C(1)'C(1)]^{-1} = D'^{-1}B' U = C sqrt[1+C'C]^{-1} , hence
>


-- Marc Nardmann

Marc Nardmann

I wrote:

>Now we show that for every p x q matrix F (e.g. F=B(t)), the matrix 1-
>F(1+F'F)^{-1}F' is positive definite: Let v be in R^p. Then v = Fx +w
>for some x in R^q and some w in R^p which is orthogonal to the image of
>F. Let G denote the positive semidefinite matrix F'F. We have to show that
>
><v,v> - <F(1+G)^{-1}F'v,v>
>
>is positive if v is nonzero. Since F'w = 0, we obtain
>
><v,v> - <F(1+G)^{-1}F'v,v> = <w,w> + <Fx,Fx> - <F(1+G)^{-1}F'Fx, Fx> ,
>
>so it suffices to prove that <Gx,x> - <(1+G)^{-1}Gx, Gx> is positive if
>x is nonzero. Let x be nonzero, y = (1+G)^{-1}x. Then
>
>0 < <Gy, y> + <GGy, y>
> = <G(1+G)y, y>
> = <Gx, (1+G)^{-1}x>
> = <Gx, (1+G)^{-1}(1+G)x -(1+G)^{-1}Gx>
> = <Gx, x> - <Gx, (1+G)^{-1}Gx> ,
>
>as claimed. Thus 1- F(1+F'F)^{-1}F' is indeed positive definite.
>
>

Oops, not quite. What I should have said is this:

> [...] so it suffices to prove that <Gx,x> - <(1+G)^{-1}Gx, Gx> is
> positive if
> Fx is nonzero. Let Fx be nonzero, y = (1+G)^{-1}x. Then Gx is nonzero
> because <Gx,x> = <Fx,Fx> is nonzero. Hence
>
> <Gy,Gx> = <G(1+G)^{-1}x, Gx> = <(1+G)^{-1}Gx,Gx> > 0 ,
>
> so Gy is nonzero. Thus
>
> 0 < <Gy,y> + <Gy,Gy> = <Gy, y> + <GGy, y>
>
> [...]


Near the end of the message, there was a typo:

>C(1) sqrt[1+C(1)'C(1)]^{-1} = D'^{-1}B' U = C sqrt[1+C'C] , hence
>

should be:

>C(1) sqrt[1+C(1)'C(1)]^{-1} = D'^{-1}B' U = C sqrt[1+C'C]^{-1} , hence
>


-- Marc Nardmann