The Submodule of Invariants
If 
is a module of a Lie algebra 
, there is one submodule that turns out to be rather interesting: the submodule 
of vectors 
such that 
for all 
. We call these vectors “invariants” of .
As an illustration of how interesting these are, consider the modules we looked at last time. What are the invariant linear maps 
from one module to another 
? We consider the action of on a linear map 
:
=x\cdot f(V)-f(x\cdot v)=0](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cleft%5Bx%5Ccdot+f%5Cright%5D%28v%29%3Dx%5Ccdot+f%28V%29-f%28x%5Ccdot+v%29%3D0&bg=e6e6e6&fg=333333&s=0 "\displaystyle\left[x\cdot f\right](v)=x\cdot f(V)-f(x\cdot v)=0")
Or, in other words:
That is, a linear map 
is invariant if and only if it intertwines the actions on and . That is, 
.
Next, consider the bilinear forms on . Here we calculate
&=-B([y,x],z)-B(x,[y,z])\\&=B([x,y],z)-B(x,[y,z])=0\end{aligned}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7D%5Cleft%5By%5Ccdot+B%5Cright%5D%28x%2Cz%29%26%3D-B%28%5By%2Cx%5D%2Cz%29-B%28x%2C%5By%2Cz%5D%29%5C%5C%26%3DB%28%5Bx%2Cy%5D%2Cz%29-B%28x%2C%5By%2Cz%5D%29%3D0%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0 "\displaystyle\begin{aligned}\left[y\cdot B\right](x,z)&=-B([y,x],z)-B(x,[y,z])\\&=B([x,y],z)-B(x,[y,z])=0\end{aligned}")
That is, a bilinear form is invariant if and only if it is associative, in the sense that the Killing form is: ![B([x,y],z)=B(x,[y,z])](http://s0.wp.com/latex.php?latex=B%28%5Bx%2Cy%5D%2Cz%29%3DB%28x%2C%5By%2Cz%5D%29&bg=e6e6e6&fg=333333&s=0 "B([x,y],z)=B(x,[y,z])")
Like this:
Like Loading...
September 21, 2012 -
Posted by John Armstrong |
Algebra, Lie Algebras, Representation Theory
Hi John, don’t know whether it’s worth posting here but your blog has been mentioned in the list of mathematics blogs here-> http://www.talkora.com/science/List-of-mathematics-blogs_112 (look for entry #5 in the list)
Will you be posting anymore for the rest of time?
Cheers,
NS