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2d
comment Book which covers these contents of the same level of Fulton's book.
If you understand the basic idea first then any differences in language (which I think are actually fairly minor) won't be a problem. If you don't understand the basic idea first, I don't think any amount of comparing the formal proofs in different books is going to matter.
2d
answered Book which covers these contents of the same level of Fulton's book.
2d
answered under change of coordinates the variety $Z(H_1,..,H_r)$ becomes $Z(x_1,…,x_r)\subset \mathbb{P}^n$.
2d
answered Group theory argument
2d
comment Irreducible $ \mathbb{C} $-character of a finite groups
Notice that we use the finiteness of $G$ in the last line: the only matrix of finite multiplicative order whose (generalized) eigenvalues are all 1 is the identity, but this doesn't hold for arbitrary matrices.
2d
answered Complete misunderstanding of Lie groups and representations
May
19
revised $ \mathbb{C} $-character table of $ D_{14} $
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May
18
comment Complete misunderstanding of Lie groups and representations
What do you mean by the direct sum of matrices?
May
18
answered $ \mathbb{C} $-character table of $ D_{14} $
May
15
comment Theorem** on page 288 from Geometry of algebraic curves by Arbarello, Cornalba, Griffiths, Harris
Yes, all nondegenerate quadric forms are equivalent.
May
15
comment Theorem** on page 288 from Geometry of algebraic curves by Arbarello, Cornalba, Griffiths, Harris
Dimension is an intrinsic property of a variety, so $\Sigma$ will be one-dimensional no matter what it's embedded in. Moreover, the Grassmannian $Gr(2,4)$, which is four-dimensional, doesn't embed into $\mathbb{P}V \cong \mathbb{P}^3$, it embeds into $\mathbb{P}(V \wedge V) \cong \mathbb{P}^5$.
May
15
comment Theorem** on page 288 from Geometry of algebraic curves by Arbarello, Cornalba, Griffiths, Harris
$\Sigma$ is a subvariety of Gr(2, V), not of $\mathbb{P}V$.
May
15
comment Theorem** on page 288 from Geometry of algebraic curves by Arbarello, Cornalba, Griffiths, Harris
It's the projective variety cut out of $\mathbb{P}V$ by the equation $x_0 x_3 + x_1 x_2 = 0$. Are you clear on how projective space works?
May
15
comment Theorem** on page 288 from Geometry of algebraic curves by Arbarello, Cornalba, Griffiths, Harris
In this case, $\Sigma$ is the set of all lines on the quadric. Each ruling is a one-dimensional family of lines, so $\Sigma$ consists of two disjoint one-dimensional components. (In fact, each component is isomorphic to $\mathbb{P}^1$.)
May
15
comment Expansion of reciprocal of quadratic
Oh, I wasn't offended! I just thought it was an incredible coincidence.
May
15
comment Expansion of reciprocal of quadratic
@AndréNicolas: You commented literally seconds before I hit "submit" on the update to my answer, in which I said just that!
May
15
revised Expansion of reciprocal of quadratic
added 526 characters in body
May
15
comment Expansion of reciprocal of quadratic
Sorry, that was just a typo.
May
15
revised Expansion of reciprocal of quadratic
added 1 character in body
May
15
answered Expansion of reciprocal of quadratic