Beach Money: Creating Your Dream Life Through Network Marketing Download Pdf [Extra Quality] ⊳
Beach Money: Creating Your Dream Life Through Network Marketing Download Pdf [Extra Quality] ⊳
Beach Money: Creating Your Dream Life Through Network Marketing Download Pdf
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Beach Money shows you that the freedom you desire is closer than you have .Q:
Why does $P$ happen to be a smooth surface?
I have problems to understand the question in the title:
Why does the parabolic subgroup $P \subset G$ of a linear algebraic group $G$ have to be smooth? If $G$ is semisimple, then we can take the connected component of $P$, and it is smooth. The key in the proof is that $P \times P$ has finitely many components and that one of them is $P \times P$.
I also assume that $P$ is a subgroup of $G$.
A:
I think the true source of the problem is that most proofs of this fact about the F-radical of a linear algebraic group begin by assuming that the group $G$ is reductive, although you do not say that. The idea is that an F-radical of a reductive group $G$ has to be a smooth irreducible subvariety of $G$ and that an irreducible smooth subvariety of $G$ is a subgroup of $G$. The failure to prove the general case is seen from the reduction to the semisimple case: consider $GL_{2}$, with subgroup $P$ of diagonal matrices. The variety $P$ is irreducible, but it is not a subgroup of $G$. (It is isomorphic to $P$ by the adjoint map.)
The semisimple case is treated first in several textbooks because it
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