We can now define a symmetric bilinear form
on our Lie algebra 
by the formula
It’s symmetric because the cyclic property of the trace lets us swap
and 
and get the same value. It also satisfies another identity which is referred to as “associativity”, though it may not appear like the familiar version of that property at first:![\displaystyle\begin{aligned}\kappa([x,y],z)&=\mathrm{Tr}(\mathrm{ad}([x,y])\mathrm{ad}(z))\\&=\mathrm{Tr}([\mathrm{ad}(x),\mathrm{ad}(y)]\mathrm{ad}(z))\\&=\mathrm{Tr}(\mathrm{ad}(x)[\mathrm{ad}(y),\mathrm{ad}(z)])\\&=\mathrm{Tr}(\mathrm{ad}(x)\mathrm{ad}([y,z]))\\&=\kappa(x,[y,z])\end{aligned}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7D%5Ckappa%28%5Bx%2Cy%5D%2Cz%29%26%3D%5Cmathrm%7BTr%7D%28%5Cmathrm%7Bad%7D%28%5Bx%2Cy%5D%29%5Cmathrm%7Bad%7D%28z%29%29%5C%5C%26%3D%5Cmathrm%7BTr%7D%28%5B%5Cmathrm%7Bad%7D%28x%29%2C%5Cmathrm%7Bad%7D%28y%29%5D%5Cmathrm%7Bad%7D%28z%29%29%5C%5C%26%3D%5Cmathrm%7BTr%7D%28%5Cmathrm%7Bad%7D%28x%29%5B%5Cmathrm%7Bad%7D%28y%29%2C%5Cmathrm%7Bad%7D%28z%29%5D%29%5C%5C%26%3D%5Cmathrm%7BTr%7D%28%5Cmathrm%7Bad%7D%28x%29%5Cmathrm%7Bad%7D%28%5By%2Cz%5D%29%29%5C%5C%26%3D%5Ckappa%28x%2C%5By%2Cz%5D%29%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0 "\displaystyle\begin{aligned}\kappa([x,y],z)&=\mathrm{Tr}(\mathrm{ad}([x,y])\mathrm{ad}(z))\\&=\mathrm{Tr}([\mathrm{ad}(x),\mathrm{ad}(y)]\mathrm{ad}(z))\\&=\mathrm{Tr}(\mathrm{ad}(x)[\mathrm{ad}(y),\mathrm{ad}(z)])\\&=\mathrm{Tr}(\mathrm{ad}(x)\mathrm{ad}([y,z]))\\&=\kappa(x,[y,z])\end{aligned}")
Where we have used the trace identity from last time.
This is called the Killing form, named for Wilhelm Killing and not nearly so coincidentally as the Poynting vector. It will be very useful to study the structures of Lie algebras.
First, though, we want to show that the definition is well-behaved. Specifically, if
is an ideal, then we can define 
to be the Killing form of 
. It turns out that is just the same as , but restricted to take its arguments in instead of all of .A lemma: if
is any subspace of a vector space and 
has its image contained in 
, then the trace of 
over 
is the same as its trace over . Indeed, take any basis of and extend it to one of ; the matrix of with respect to this basis has zeroes for all the rows that do not correspond to the basis of , so the trace may as well just be taken over .Now the fact that
is an ideal means that for any 
the mapping 
is an endomorphism of sending all of into . Thus its trace over is the same as its trace over all of , and the Killing form on applied to is the same as the Killing form on applied to the same two elements.
DIGITAL JUICE
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Thank's!