Undergrad What does V^G mean in linear algebra?

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The notation V^G in linear algebra refers to the set of elements in the vector space V that are invariant under the action of the group G. This means that V^G consists of those elements that remain unchanged when any group element acts on them. The discussion clarifies that while X^Y typically denotes functions from Y to X, in this context, it specifically pertains to invariant elements under group action. It also emphasizes that the AI summary incorrectly conflates V^G with V^*, leading to misleading conclusions. Understanding V^G is crucial for studying linear algebraic groups and their properties.
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Hi. Newbee question. What does the notation V^G mean where V is a vector space and G is a group? I found it in: A linear algebraic group G is called linearly reductive if for every rational representation V and every v in V^G \ {0}, there exists a linear invariant function f in (V^*)^G such that f(v)<>0. I can't find a definition of the notation using google. Thanks.
Hi. Newbee question. What does the notation V^G mean where V is a vector space and G is a group? I found it in: A linear algebraic group G is called linearly reductive if for every rational representation V and every v in V^G \ {0}, there exists a linear invariant function f in (V^*)^G such that f(v)<>0. I can't find a definition of the notation using google. Thanks.
 
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Normally for sets ##X## and ##Y##, ##X^Y## is the set of functions from ##Y## to ##X##. In a context like this, it's often meant to be limited to functions of the desired type, e.g. only representations.

That doesn't really make sense here, since a representation is a map ##G\to GL(V)##, not a map from ##G## to ##V##.
 
When a group ##G## acts on a set ##X## (in your case ##V##), sometimes one writes ##X^G## for the set of invariant elements, that is, the elements of ##X## that are fixed by all group elements. Note that if ##G## acts linearly on ##V##, then it also does on ##V^*##, so it makes sense to write ##(V^*)^G## in your example.
 
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The AI summary explanation that ends this thread, visible only when not logged in, incorrectly equates V^G with V ^*, and hence virtually every statement in it is false.
 
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