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 a vector space V that remain invariant under the action of a group G. Specifically, when G acts on V, V^G denotes the subset of V that is fixed by all elements of G. This concept is crucial in the context of linear algebraic groups, particularly when discussing linearly reductive groups and their representations. Misinterpretations, such as equating V^G with V^*, are incorrect and lead to false conclusions.

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Mathematicians, students of linear algebra, and researchers in algebraic geometry or representation theory will benefit from this discussion, particularly those interested in the intersection of group theory and linear algebra.

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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.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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