Soergel diagrammatics for dihedral groups
... The Temperley-Lieb category can be defined over Z[q + q −1 ]. Every quantum number in q can be expressed as a polynomial in [2] = q + q −1 . We write ζm for an arbitrary primitive m-th root of unity. The statement that q 2 = ζm is equivalent to the statement that [m] = 0 and [n] 6= 0 for n < m. So w ...
... The Temperley-Lieb category can be defined over Z[q + q −1 ]. Every quantum number in q can be expressed as a polynomial in [2] = q + q −1 . We write ζm for an arbitrary primitive m-th root of unity. The statement that q 2 = ζm is equivalent to the statement that [m] = 0 and [n] 6= 0 for n < m. So w ...
Algebraic algorithms Freely using the textbook: Victor Shoup’s “A Computational P´eter G´acs
... would be different elemnts x, y with x · b = y · b, but (x − y) · b 6= 0, since b is not a zero divisor. At the beginning of class, we have seen that in a finite set, if a class is one-to-one then it is also onto. Therefore for each c there is an x with x · b = c. The one-to-on ...
... would be different elemnts x, y with x · b = y · b, but (x − y) · b 6= 0, since b is not a zero divisor. At the beginning of class, we have seen that in a finite set, if a class is one-to-one then it is also onto. Therefore for each c there is an x with x · b = c. The one-to-on ...
13. Dedekind Domains
... R be the integral closure of Z in K. Then R is a Dedekind domain. Proof. As a subring of a field, R is clearly an integral domain. Moreover, by Example 11.3 (c) and Lemma 11.8 we have dim R = dim Z = 1. It is also easy to see that R is normal: if a ∈ Quot R ⊂ K is integral over R it is also integral ...
... R be the integral closure of Z in K. Then R is a Dedekind domain. Proof. As a subring of a field, R is clearly an integral domain. Moreover, by Example 11.3 (c) and Lemma 11.8 we have dim R = dim Z = 1. It is also easy to see that R is normal: if a ∈ Quot R ⊂ K is integral over R it is also integral ...
Cohomology and K-theory of Compact Lie Groups
... By Lemma 2.1, Ω∗L (G) is a complex with differential d. Let HL∗ (G) be the cohomology of Ω∗L (G). Lemma 2.2 implies that HL∗ (G) has a ring structure induced by wedge product. Since Ω∗L (G) is a subspace of Ω∗ (G), there is an inclusion map ι : Ω∗L (G) → Ω∗ (G), which induces ι∗ : HL∗ (G) → H ∗ (G, ...
... By Lemma 2.1, Ω∗L (G) is a complex with differential d. Let HL∗ (G) be the cohomology of Ω∗L (G). Lemma 2.2 implies that HL∗ (G) has a ring structure induced by wedge product. Since Ω∗L (G) is a subspace of Ω∗ (G), there is an inclusion map ι : Ω∗L (G) → Ω∗ (G), which induces ι∗ : HL∗ (G) → H ∗ (G, ...
SOLVING a ± b = 2c IN THE ELEMENTS OF FINITE SETS 1
... Analyzing the expression in the right-hand side we see that if (13) were false, then N < O along with (9) would give x0 + x1 − y0 − y1 < x0 − y1 − (1 − s), which is (13) in disguise. This contradiction shows that (13) is true. We now readily get (12) as a consequence of (13) and (9), and (14) is jus ...
... Analyzing the expression in the right-hand side we see that if (13) were false, then N < O along with (9) would give x0 + x1 − y0 − y1 < x0 − y1 − (1 − s), which is (13) in disguise. This contradiction shows that (13) is true. We now readily get (12) as a consequence of (13) and (9), and (14) is jus ...
Whitney forms of higher degree
... for nodes (0-simplices), edges (1-simplices), etc., each with its own orientation. Note that e (resp., f , v) is by definition an ordered couple (resp., triplet, quadruplet) of vertices, not merely a collection. For example, the edge e = {, n} is oriented from the node to n. Given a domain Ω ⊂ Rd ...
... for nodes (0-simplices), edges (1-simplices), etc., each with its own orientation. Note that e (resp., f , v) is by definition an ordered couple (resp., triplet, quadruplet) of vertices, not merely a collection. For example, the edge e = {, n} is oriented from the node to n. Given a domain Ω ⊂ Rd ...
Script 2013W 104.271 Discrete Mathematics VO (Gittenberger)
... • The tree with two nodes: in this case there is one edge, connecting the two leaves. • A tree T , with at least three nodes: start at an arbitrary node, this node has to have a neighbor: – If the node has only one neighbor, remove the edge and this node, this gives a new tree: T 0 . The last part o ...
... • The tree with two nodes: in this case there is one edge, connecting the two leaves. • A tree T , with at least three nodes: start at an arbitrary node, this node has to have a neighbor: – If the node has only one neighbor, remove the edge and this node, this gives a new tree: T 0 . The last part o ...
Why eigenvalue problems?
... Now suppose we choose some fixed λ ∈ C and look at the 2-by-2 matrices for which λ is an eigenvalue. If we just want λ to be an eigenvalue, we must satisfy one scalar equation: p(λ) = 0. To find matrices for which λ is a double eigenvalue, we must satisfy the additional constraint a + d = 2λ. And th ...
... Now suppose we choose some fixed λ ∈ C and look at the 2-by-2 matrices for which λ is an eigenvalue. If we just want λ to be an eigenvalue, we must satisfy one scalar equation: p(λ) = 0. To find matrices for which λ is a double eigenvalue, we must satisfy the additional constraint a + d = 2λ. And th ...
Introduction to amoebas and tropical geometry
... j,k:j+k=i aj bk . If ci 6= 0 then there are j, k ∈ Z such that aj 6= 0 and bk 6= 0, so i = j + k with j ∈ Newt(f ) and k ∈ Newt(g). Thus Newt(f g) ⊂ Newt(f ) + Newt(g). To show the converse it is enough to show that all vertices of Newt(f ) + Newt(g) are contained in Newt(f g). If k is a vertex of N ...
... j,k:j+k=i aj bk . If ci 6= 0 then there are j, k ∈ Z such that aj 6= 0 and bk 6= 0, so i = j + k with j ∈ Newt(f ) and k ∈ Newt(g). Thus Newt(f g) ⊂ Newt(f ) + Newt(g). To show the converse it is enough to show that all vertices of Newt(f ) + Newt(g) are contained in Newt(f g). If k is a vertex of N ...
