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A complete set of centrally primitive idempotents $\{e_1,\ldots, e_n\}$ in $R$ furnishes a decomposition of of $R$ into directly indecomposable rings, namely the indecomposable rings $e_iRe_i$. A ring may have a complete set of centrally primitive idempotents without having a complete set of primitive idempotents. These idempotents do not connect well with ...


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The action on morphisms is given by restricting a morphism to invariants. There's nothing surprising going on here. $V^G$ is representable: in fact it is represented by $k$ (the base field) with the trivial action of $G$. Hence it is not only left exact but continuous, and in fact has a left adjoint. $k$ being algebraically closed is irrelevant. Your ...


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Let the value on each of the six sides of the dice be $x_1, x_2, ... , x_6$ You can create equations for each new value. So for example, the value $x_1$ will be replaced by the average of $x_2, x_3, x_4$ and $x_5$ We can say that $x_{1,n+1} = \frac 1 4 x_{2,n} + \frac 1 4 x_{3,n}+ \frac 1 4 x_{4,n}+ \frac 1 4 x_{5,n} $ You now need to write down the other ...


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I don't know how to bring representation theory into the game. But anyway: Here is a solution proposal that simplifies and generalizes this problem at the same time. Write any real or complex numbers you like on the faces of the cube $[{-1},1]^3$. Denote the average of the two numbers on the opposite faces $x_i=\pm 1$ by $z_i$, whereby $i$ is taken modulo ...


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All $S_n$'s have an irreducible representation of degree $n-1$: take the natural permutation representation of degree $n$, say on the permuted basis $v_1,v_2,\ldots,v_n$, then the subspace spanned by $v_j-v_{j-1},\ j>1$ is in fact a subrepresentation and is irreducible. You can always multiply this representation by signum and get another irreducible ...


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I'm very sorry for the late reply but yes, it is true and it follows from the definition. Indeed, if $f(\cdot) = \left\langle \phi(\cdot)v, w\right\rangle$ is a matrix coefficient of $G$, and $g$ is any (continuous) function on $G$, then: $$(f\ast g)(x)$$ $$=\int_{G} f(xh^{-1})g(h) dh$$ $$=\int_{G} \left\langle \phi(xh^{-1})v,w\right\rangle g(h)dh$$ ...


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The sum is just over a single element: $$\langle\mathbb I, \mathrm{Res}_{\{I\}}\chi\rangle_{\{I\}} = \frac{1}{1}\sum_{g \in \{I\}}\chi(g)\overline{\mathbb I(g)} = \chi(I)\overline{\mathbb I(I)} = \dim\chi\cdot\dim\mathbb I = \dim\chi$$ Now observe that characters form an inner product space with an orthogonal basis. That means characters are defined by ...



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