# Read e-book online A basis for the right quantum algebra and the “1 = q” PDF

By Foata D., Han G.-N.

**Read or Download A basis for the right quantum algebra and the “1 = q” principle PDF**

**Similar algebra books**

- Übungsaufgaben zur linearen Algebra und linearen Optimierung
- Algebra für Informatiker
- Topics in Cohomological Studies of Algebraic Varieties: Impanga Lecture Notes
- Algebra
- Groupes de Galois arithmétiques et différentiels

**Extra info for A basis for the right quantum algebra and the “1 = q” principle**

**Example text**

Some examples. 1. 5) p ∈ Rn → (p, v) is a vector field. Vector fields of this type are constant vector fields. 2. In particular let ei , i = 1, . . , n, be the standard basis vectors of Rn . 5) by ∂/∂x i . 1 Vector fields and one-forms 51 3. Given a vector field on U and a function, f : U → R we’ll denote by f v the vector field p ∈ U → f (p)v(p) . 4. Given vector fields v1 and v2 on U , we’ll denote by v1 + v2 the vector field p ∈ U → v1 (p) + v2 (p) . 5. The vectors, (p, ei ), i = 1, . . , n, are a basis of Tp Rn , so if v is a vector field on U , v(p) can be written uniquely as a linear combination of these vectors with real numbers, g i (p), i = 1, .

C ,τ ( ) where σ ∈ Sk and τ ∈ S . 8. Let V and W be vector spaces and let A : V → W be a linear map. Show that if Av = w then for ω ∈ Λp (w∗ ), A∗ ι(w)ω = ι(v)A∗ ω . ) 44 Chapter 1. 9 Orientations We recall from freshman calculus that if ⊆ R 2 is a line through the origin, then −{0} has two connected components and an orientation of is a choice of one of these components (as in the figure below). ✣ ✡ ✡ •✡0 ✡ More generally, if L is a one-dimensional vector space then L−{0} consists of two components: namely if v is an element of L−[0}, then these two components are and L1 = {λv λ > 0} L2 = {λv, λ < 0} .

Therefore since π(W ) = 0 π(T σ ) = (−1)σ π(T ) . 10) 1 and 1 2 ∧ Tσ = ∧ ··· ∧ = (−1)σ In particular, for k, 1 σ(k) ∧ ··· ∧ σ(1) =− 2 ∧ σ(k) , = (−1)σ π(T ) k ∈V∗ 2 ⊗ ··· ⊗ 1 . 11) 2 and 3 1 ∧ 33 ∈V∗ 2 ∧ 3 =− 2 ∧ 1 ∧ 3 = 2 ∧ 3 ∧ 1 . 8) the following result (which we’ll leave as an exercise). 1. If ω1 ∈ Λr and ω2 ∈ Λs then ω1 ∧ ω2 = (−1)rs ω2 ∧ ω1 . , for ω1 = 1 ∧ · · · ∧ r and ω2 = 1 ∧ · · · ∧ s . 10). Let e1 , . . , en be a basis of V and let e∗1 , . . , e∗n be the dual basis of V ∗ . 13) e∗i1 ∧ · · · e∗ik = π(e∗I ) = π(e∗i1 ⊗ · · · ⊗ e∗ik ) .

### A basis for the right quantum algebra and the “1 = q” principle by Foata D., Han G.-N.

by Charles

4.3