How to show bilinearity
Given a Coxeter group $varGamma =langle rho_0, rho_1, ldots, rho_{n - 1}rangle$ which at least satisfy the relation $(rho_irho_j)^{p_{ij}}=1, 0leq i, j leq n - 1,$ where $p_{ii}=1$ and $2 leq p_{ij}leq infty$ for all $ineq j.$ On an n - dimensional real vector space V, with basis $a_0, a_1, ldots, a_{n - 1},$ we define a symmetric bilinear form $xcdot y$ by setting $a_i cdot a_j:=-2 text{cos} frac{pi}{p_{ij}}, 0leq i, j, leq n - 1.$
How do I show bilinearity? Thanks
vector-spaces bilinear-form coxeter-groups
add a comment |
Given a Coxeter group $varGamma =langle rho_0, rho_1, ldots, rho_{n - 1}rangle$ which at least satisfy the relation $(rho_irho_j)^{p_{ij}}=1, 0leq i, j leq n - 1,$ where $p_{ii}=1$ and $2 leq p_{ij}leq infty$ for all $ineq j.$ On an n - dimensional real vector space V, with basis $a_0, a_1, ldots, a_{n - 1},$ we define a symmetric bilinear form $xcdot y$ by setting $a_i cdot a_j:=-2 text{cos} frac{pi}{p_{ij}}, 0leq i, j, leq n - 1.$
How do I show bilinearity? Thanks
vector-spaces bilinear-form coxeter-groups
I don't think you have to show bilinearity - you are told that $xcdot y$ is bilinear. Together with the $n^2$ values of $a_i cdot a_j$ this defines the value of $x cdot y$ for any $x$ and $y$. If you did not know the function was bilinear then you would know nothing about its value for general $x$ and $y$.
– gandalf61
Nov 30 '18 at 13:14
@gandalf61 I just want to verify how's it bilinear? Thanks
– primer
Nov 30 '18 at 15:21
You aren't given enough information to verify that it is bilinear. You are just told it is bilinear. It's like this. Suppose you start a proof with "Let $p$ be a prime number ..." and someone asks "But how can I verify that $p$ is a prime number ?" ... they can't, they just have to take it as given that $p$ is a prime number and carry on from there.
– gandalf61
Nov 30 '18 at 15:40
If you are actually asking "what does bilinear mean ?", see en.wikipedia.org/wiki/Bilinear_form
– gandalf61
Nov 30 '18 at 15:44
@gandalf61 I was able to read about bilinear forms. I just don't know how to verify linearity in the above case. Thanks
– primer
Nov 30 '18 at 16:19
add a comment |
Given a Coxeter group $varGamma =langle rho_0, rho_1, ldots, rho_{n - 1}rangle$ which at least satisfy the relation $(rho_irho_j)^{p_{ij}}=1, 0leq i, j leq n - 1,$ where $p_{ii}=1$ and $2 leq p_{ij}leq infty$ for all $ineq j.$ On an n - dimensional real vector space V, with basis $a_0, a_1, ldots, a_{n - 1},$ we define a symmetric bilinear form $xcdot y$ by setting $a_i cdot a_j:=-2 text{cos} frac{pi}{p_{ij}}, 0leq i, j, leq n - 1.$
How do I show bilinearity? Thanks
vector-spaces bilinear-form coxeter-groups
Given a Coxeter group $varGamma =langle rho_0, rho_1, ldots, rho_{n - 1}rangle$ which at least satisfy the relation $(rho_irho_j)^{p_{ij}}=1, 0leq i, j leq n - 1,$ where $p_{ii}=1$ and $2 leq p_{ij}leq infty$ for all $ineq j.$ On an n - dimensional real vector space V, with basis $a_0, a_1, ldots, a_{n - 1},$ we define a symmetric bilinear form $xcdot y$ by setting $a_i cdot a_j:=-2 text{cos} frac{pi}{p_{ij}}, 0leq i, j, leq n - 1.$
How do I show bilinearity? Thanks
vector-spaces bilinear-form coxeter-groups
vector-spaces bilinear-form coxeter-groups
edited Nov 30 '18 at 12:56
primer
asked Nov 30 '18 at 12:12
primerprimer
1
1
I don't think you have to show bilinearity - you are told that $xcdot y$ is bilinear. Together with the $n^2$ values of $a_i cdot a_j$ this defines the value of $x cdot y$ for any $x$ and $y$. If you did not know the function was bilinear then you would know nothing about its value for general $x$ and $y$.
– gandalf61
Nov 30 '18 at 13:14
@gandalf61 I just want to verify how's it bilinear? Thanks
– primer
Nov 30 '18 at 15:21
You aren't given enough information to verify that it is bilinear. You are just told it is bilinear. It's like this. Suppose you start a proof with "Let $p$ be a prime number ..." and someone asks "But how can I verify that $p$ is a prime number ?" ... they can't, they just have to take it as given that $p$ is a prime number and carry on from there.
– gandalf61
Nov 30 '18 at 15:40
If you are actually asking "what does bilinear mean ?", see en.wikipedia.org/wiki/Bilinear_form
– gandalf61
Nov 30 '18 at 15:44
@gandalf61 I was able to read about bilinear forms. I just don't know how to verify linearity in the above case. Thanks
– primer
Nov 30 '18 at 16:19
add a comment |
I don't think you have to show bilinearity - you are told that $xcdot y$ is bilinear. Together with the $n^2$ values of $a_i cdot a_j$ this defines the value of $x cdot y$ for any $x$ and $y$. If you did not know the function was bilinear then you would know nothing about its value for general $x$ and $y$.
– gandalf61
Nov 30 '18 at 13:14
@gandalf61 I just want to verify how's it bilinear? Thanks
– primer
Nov 30 '18 at 15:21
You aren't given enough information to verify that it is bilinear. You are just told it is bilinear. It's like this. Suppose you start a proof with "Let $p$ be a prime number ..." and someone asks "But how can I verify that $p$ is a prime number ?" ... they can't, they just have to take it as given that $p$ is a prime number and carry on from there.
– gandalf61
Nov 30 '18 at 15:40
If you are actually asking "what does bilinear mean ?", see en.wikipedia.org/wiki/Bilinear_form
– gandalf61
Nov 30 '18 at 15:44
@gandalf61 I was able to read about bilinear forms. I just don't know how to verify linearity in the above case. Thanks
– primer
Nov 30 '18 at 16:19
I don't think you have to show bilinearity - you are told that $xcdot y$ is bilinear. Together with the $n^2$ values of $a_i cdot a_j$ this defines the value of $x cdot y$ for any $x$ and $y$. If you did not know the function was bilinear then you would know nothing about its value for general $x$ and $y$.
– gandalf61
Nov 30 '18 at 13:14
I don't think you have to show bilinearity - you are told that $xcdot y$ is bilinear. Together with the $n^2$ values of $a_i cdot a_j$ this defines the value of $x cdot y$ for any $x$ and $y$. If you did not know the function was bilinear then you would know nothing about its value for general $x$ and $y$.
– gandalf61
Nov 30 '18 at 13:14
@gandalf61 I just want to verify how's it bilinear? Thanks
– primer
Nov 30 '18 at 15:21
@gandalf61 I just want to verify how's it bilinear? Thanks
– primer
Nov 30 '18 at 15:21
You aren't given enough information to verify that it is bilinear. You are just told it is bilinear. It's like this. Suppose you start a proof with "Let $p$ be a prime number ..." and someone asks "But how can I verify that $p$ is a prime number ?" ... they can't, they just have to take it as given that $p$ is a prime number and carry on from there.
– gandalf61
Nov 30 '18 at 15:40
You aren't given enough information to verify that it is bilinear. You are just told it is bilinear. It's like this. Suppose you start a proof with "Let $p$ be a prime number ..." and someone asks "But how can I verify that $p$ is a prime number ?" ... they can't, they just have to take it as given that $p$ is a prime number and carry on from there.
– gandalf61
Nov 30 '18 at 15:40
If you are actually asking "what does bilinear mean ?", see en.wikipedia.org/wiki/Bilinear_form
– gandalf61
Nov 30 '18 at 15:44
If you are actually asking "what does bilinear mean ?", see en.wikipedia.org/wiki/Bilinear_form
– gandalf61
Nov 30 '18 at 15:44
@gandalf61 I was able to read about bilinear forms. I just don't know how to verify linearity in the above case. Thanks
– primer
Nov 30 '18 at 16:19
@gandalf61 I was able to read about bilinear forms. I just don't know how to verify linearity in the above case. Thanks
– primer
Nov 30 '18 at 16:19
add a comment |
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I don't think you have to show bilinearity - you are told that $xcdot y$ is bilinear. Together with the $n^2$ values of $a_i cdot a_j$ this defines the value of $x cdot y$ for any $x$ and $y$. If you did not know the function was bilinear then you would know nothing about its value for general $x$ and $y$.
– gandalf61
Nov 30 '18 at 13:14
@gandalf61 I just want to verify how's it bilinear? Thanks
– primer
Nov 30 '18 at 15:21
You aren't given enough information to verify that it is bilinear. You are just told it is bilinear. It's like this. Suppose you start a proof with "Let $p$ be a prime number ..." and someone asks "But how can I verify that $p$ is a prime number ?" ... they can't, they just have to take it as given that $p$ is a prime number and carry on from there.
– gandalf61
Nov 30 '18 at 15:40
If you are actually asking "what does bilinear mean ?", see en.wikipedia.org/wiki/Bilinear_form
– gandalf61
Nov 30 '18 at 15:44
@gandalf61 I was able to read about bilinear forms. I just don't know how to verify linearity in the above case. Thanks
– primer
Nov 30 '18 at 16:19