GROUPS WITH FEW CONJUGACY CLASSES 1. Introduction
... classes each of which has length at most |G|. Thus pn = |N | ≤ 2 p|G|, which implies the above inequality. Step 7: N cannot be an imprimitive module over GF (p)G. Suppose that N is an imprimitive module over GF (p)G. Then N = N1 ×. . .×Nr , where the Ni ’s are permuted by G. Let r be as large as pos ...
... classes each of which has length at most |G|. Thus pn = |N | ≤ 2 p|G|, which implies the above inequality. Step 7: N cannot be an imprimitive module over GF (p)G. Suppose that N is an imprimitive module over GF (p)G. Then N = N1 ×. . .×Nr , where the Ni ’s are permuted by G. Let r be as large as pos ...
Complex numbers
... below–don’t peek!) You try (2): If z = 5 + i and w = 2 + 2i, what is w + z? zw? (answer to be given in class) We can also introduce division of complex numbers, but for that we need a couple more concepts. ...
... below–don’t peek!) You try (2): If z = 5 + i and w = 2 + 2i, what is w + z? zw? (answer to be given in class) We can also introduce division of complex numbers, but for that we need a couple more concepts. ...
Problem Set 3
... 2. Show that σ has order ≤ 3. Why can’t it have order 2? Deduce that the order of σ is 3. 3. Prove that Aut(K/k) = C3 ; that is, it is generated by σ (hint: K must be gotten by the Kronecker construction on k for a polynomial of degree 3). 4. Prove that if a polynomial f is irreducible in k[t] and h ...
... 2. Show that σ has order ≤ 3. Why can’t it have order 2? Deduce that the order of σ is 3. 3. Prove that Aut(K/k) = C3 ; that is, it is generated by σ (hint: K must be gotten by the Kronecker construction on k for a polynomial of degree 3). 4. Prove that if a polynomial f is irreducible in k[t] and h ...
The equivariant spectral sequence and cohomology with local coefficients Alexander I. Suciu
... is trivial for all j ≤ q if and only if H (H (X, k), ·νk ) = 0, for all j ≤ q. Particularly interesting is the case of a smooth manifold X fibering over the circle, with ν = p∗ : π Z the homomorphism induced by the projection map, p : X → S 1 . The homology of the resulting infinite cyclic cover w ...
... is trivial for all j ≤ q if and only if H (H (X, k), ·νk ) = 0, for all j ≤ q. Particularly interesting is the case of a smooth manifold X fibering over the circle, with ν = p∗ : π Z the homomorphism induced by the projection map, p : X → S 1 . The homology of the resulting infinite cyclic cover w ...
Homework 9 - Material from Chapters 9-10
... Solution: This question is asking how many times the coset 14 + h8i must be added to itself in order to get the identity coset 0 + h8i. To equal the identity coset we must have a + h8i where a ∈ h8i, so a = 0, 8, or 16. We can just add 14 to itself in Z24 until we get 0 8, or 16, and see how many ti ...
... Solution: This question is asking how many times the coset 14 + h8i must be added to itself in order to get the identity coset 0 + h8i. To equal the identity coset we must have a + h8i where a ∈ h8i, so a = 0, 8, or 16. We can just add 14 to itself in Z24 until we get 0 8, or 16, and see how many ti ...
Midterm solutions.
... The Bolzano Weierstrass theorem asserts that closed bounded intervals in R are sequentially compact. Assuming only this theorem, prove that closed balls in R3 are sequentially compact. Let C be a closed ball in R3 of radius r, centred at (a, b, c) and (xn , yn , zn ) be a sequence in C . Since (xn − ...
... The Bolzano Weierstrass theorem asserts that closed bounded intervals in R are sequentially compact. Assuming only this theorem, prove that closed balls in R3 are sequentially compact. Let C be a closed ball in R3 of radius r, centred at (a, b, c) and (xn , yn , zn ) be a sequence in C . Since (xn − ...
linear representations as modules for the group ring
... for µ(a, b). An identity for an associative algebra A is an element 1 of A such that 1a = a1 = a for all a in A. We can talk about units in an associative algebra with identity: a in A is a unit if there is an element b of A such that ab = ba = 1.5 Units form a group under multiplication. A map of a ...
... for µ(a, b). An identity for an associative algebra A is an element 1 of A such that 1a = a1 = a for all a in A. We can talk about units in an associative algebra with identity: a in A is a unit if there is an element b of A such that ab = ba = 1.5 Units form a group under multiplication. A map of a ...
solutions to HW#3
... given as the product of disjoint cycles of sizes 5, 2, 3, and 2. The order of σ is the least common multiple of these numbers, so |σ| = 30. 1.3.7 Write out the cycle decomposition of each element of order 2 in S4 . The six transpositions all have order 2. They are (12), (13), (14), (23), (24), and ( ...
... given as the product of disjoint cycles of sizes 5, 2, 3, and 2. The order of σ is the least common multiple of these numbers, so |σ| = 30. 1.3.7 Write out the cycle decomposition of each element of order 2 in S4 . The six transpositions all have order 2. They are (12), (13), (14), (23), (24), and ( ...
LECTURES ON SYMPLECTIC REFLECTION ALGEBRAS 2. Algebras of Crawley-Boevey and Holland
... xi y j γ, where i + j 6 n form a basis in Hc6n . Now we can take the spherical subalgebra eHc e ⊂ Hc . This algebra is filtered, we just restrict the filtration from Hc , i.e., (eHc e)6n := eHc e∩Hc6n . Equivalently, (eHc e)6n = eHc6n e. Exercise 2.8. Deduce gr eHc e = C[x, y]Γ from gr Hc = C[x, y]#Γ ...
... xi y j γ, where i + j 6 n form a basis in Hc6n . Now we can take the spherical subalgebra eHc e ⊂ Hc . This algebra is filtered, we just restrict the filtration from Hc , i.e., (eHc e)6n := eHc e∩Hc6n . Equivalently, (eHc e)6n = eHc6n e. Exercise 2.8. Deduce gr eHc e = C[x, y]Γ from gr Hc = C[x, y]#Γ ...
PDF on arxiv.org - at www.arxiv.org.
... is a differential of the first kind on C for each (j, i) ∈ Ln,p . This implies easily that the collection {ωj,i }(j,i)∈Ln,p is a basis in the space of differentials of the first kind on C. There is a non-trivial birational Ka -automorphism of C δp : (x, y) 7→ (x, ζy). Clearly, δpp is the identity ma ...
... is a differential of the first kind on C for each (j, i) ∈ Ln,p . This implies easily that the collection {ωj,i }(j,i)∈Ln,p is a basis in the space of differentials of the first kind on C. There is a non-trivial birational Ka -automorphism of C δp : (x, y) 7→ (x, ζy). Clearly, δpp is the identity ma ...
Full text in
... We prove that (Ap (ω), ∥.∥p,ω ) is Hermitian. In the particular case where F is a harmonic function in a neighborhood of f (R), we prove that the expression of F (f ) is also given by the Poisson integral formula ([1]). 2. Real analytic version of Levy’s theorem Now we are ready to generalize Levy’s ...
... We prove that (Ap (ω), ∥.∥p,ω ) is Hermitian. In the particular case where F is a harmonic function in a neighborhood of f (R), we prove that the expression of F (f ) is also given by the Poisson integral formula ([1]). 2. Real analytic version of Levy’s theorem Now we are ready to generalize Levy’s ...
here.
... (Ann M )p = Ann(Mp ) for any prime ideal p. If p ⊇ Ann M , then pAp ⊇ (Ann M )p = Ann(Mp ), hence Ann(Mp ) 6= Ap , and therefore Mp 6= 0. (Alternatively: Assume M is finitely generated, say by m1 , . . . , mn . Suppose p ⊂ A is a prime ideal such that Mp = 0. Then there are elements s1 , . . . , sn ...
... (Ann M )p = Ann(Mp ) for any prime ideal p. If p ⊇ Ann M , then pAp ⊇ (Ann M )p = Ann(Mp ), hence Ann(Mp ) 6= Ap , and therefore Mp 6= 0. (Alternatively: Assume M is finitely generated, say by m1 , . . . , mn . Suppose p ⊂ A is a prime ideal such that Mp = 0. Then there are elements s1 , . . . , sn ...
Topology Semester II, 2014–15
... (i) The ordered square is locally connected: just observe that any neighborhood U of any point x × y contains an interval of the form (a × b, c × d) for a × b < x × y < c × d by definition of the order topology. By Theorem 24.1 in the book, an (open) interval of a linear continuum is connected, so ( ...
... (i) The ordered square is locally connected: just observe that any neighborhood U of any point x × y contains an interval of the form (a × b, c × d) for a × b < x × y < c × d by definition of the order topology. By Theorem 24.1 in the book, an (open) interval of a linear continuum is connected, so ( ...
Here
... the identity. If x is an element of order pq, then xq is of order p. Therefore, if G does not contain an element of order p, then all non-identity elements are of order q. Now the cyclic groups generated by each element must either be equal, or have only the identity in common, any non-identity elem ...
... the identity. If x is an element of order pq, then xq is of order p. Therefore, if G does not contain an element of order p, then all non-identity elements are of order q. Now the cyclic groups generated by each element must either be equal, or have only the identity in common, any non-identity elem ...
Complex exponentials: Euler`s formula
... The second formula is not quite precise. The argument of a number is not unique, since we can add to it a multiple of 2π and not change the position of the point. Hence we have to interpret the second formula a bit loosely. For example, arg(−i) = 3π/2 but arg((−i)(−i)) = arg(−1) = π, altho 3π/2 + 3π ...
... The second formula is not quite precise. The argument of a number is not unique, since we can add to it a multiple of 2π and not change the position of the point. Hence we have to interpret the second formula a bit loosely. For example, arg(−i) = 3π/2 but arg((−i)(−i)) = arg(−1) = π, altho 3π/2 + 3π ...
1. Fundamental Group Let X be a topological space. A path γ on X is
... Let X be a topological space. A path γ on X is a continuous map γ : [0, 1] → X. The points γ(0) and γ(1) are called initial point and terminal point respectively. Two points p, q in X are said to be connected by a path if there is a curve γ whose initial point is p and terminal point is q. We say th ...
... Let X be a topological space. A path γ on X is a continuous map γ : [0, 1] → X. The points γ(0) and γ(1) are called initial point and terminal point respectively. Two points p, q in X are said to be connected by a path if there is a curve γ whose initial point is p and terminal point is q. We say th ...