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Superspaces Definition. the Superspace E Is Z2-Graded Vector Superspaces Definition. The superspace E is Z2-graded vector space E = E+ ⊕ E−. A superalgebra is an algebra A whose underlying vector space is a superspace, and whose products respects the Z2-grading; in other words, Ai · Aj ⊂ Ai+j . Example 1. An exterior algebra is a supertrace with the Z2-grading Λ±E = X ΛiE. (−1)i=±1 + Example 2. An ungraded vector space E is implicitly Z2-graded with E = E and E− =0. • The algebra of endomorphisms End(E) of a superspace is a superalgebra, when graded in the usual way: End+(E) =Hom(E+,E+) ⊕ Hom(E−,E−) End+(E) =Hom(E+,E+) ⊕ Hom(E−,E−). Definition. A superbundle on a manifold M is a bundle E = E+ ⊕E−, where E+ and E− are two vector bundles on M such that the fibers of E are superspaces. • If E is a vector bundle, we will identify it with the superbundle E+ = E and E− =0. • There is a general principle in dealing with elements of a superspace, that whenever a formula involves commuting an element a past another b, one must insert a sign (−1)|a||b|, where |a| is the parity of a, which equals 0 or 1 accrording to whether the degree of a is even or odd. – Thus, the supercommutator of a pair of odd parity elements of a superalgebra is actually their anticommutator, due to the extra minus sign. • We will use the same bracket notation for this supercommutator as is usually used for the commutator: [a, b]=ab − (−1)|a|·|b|ba. This bracket satisfies the axioms of a Lie superalgebra: (i) [a, b]+(−1)|a|·|b|[b, a]=0 (ii)[a, [b, c]] = [[a, b],c]+(−1)|a|·|b|[b, [a, c]]. Typeset by AMS-TEX 1 2 Definition. A superalgebra is said to be super-commutative if its superbracket vanishes identically. Example. The exterior algebra ΛE is super-commutative. Definition. A on a superalgebra A is a linear form φn A satisfying φ([a, b]) = 0. • On a superalgebra End(E), there is a canonical linear form given by the formula Tr + (a) − Tr − (a), if a is even, Str(a)= E E 0, if a is odd. Proposition 1. The linear form Str defined above is a supertrace on End(E). Proof. We must verify that Str[a, b]=0. (i) If a and b have opposite parity, then [a, b] is odd in parity and hence Str[a, b]=0. a+ 0 b+ 0 (ii) If a = and b = are both even, then 0 a− 0 b− [a+,b+]0 [a, b]= 0[a−,b−] has vanishing supertrace, since Tr[a+,b+]=Tr[a−,b−]=0. 0 a− 0 b− (iii) If a = and b = are both odd, then a+ 0 b+ 0 a−b+ + b−a+ 0 [a, b]= 0 a+b− + b+a− has supertrace − + − + + − + − Str[a, b]=TrE+ (a b + b a ) − TrE+ (a b + b a )=0. • If E = E+ ⊕ E− and F = F + ⊕ F − are two supertraces, then their tensor product E × F is the superapace with underlying vector space E ⊗ F and grading (E ⊗ F )+ =(E+ ⊗ F +) ⊕ (E− ⊗ F −), (E ⊗ F )− =(E+ ⊗ F −) ⊕ (E− ⊗ F +). • If A and B are superalgebras, then their tensor product A⊗B is the superalgebra whose underlying space is Z2-graded tensor product of A and B, and whose product is defined by the rule |b1|·|a2| (a1 ⊗ b1) · (a2 ⊗ b2)=(−1) a1a2 ⊗ b1b2. 3 Definition. If A is a supercommutative algebra and E is a supertrace, we extend the supertrace on End (E) to a map Str : A ⊗ End(E) → A, by the formula Str(a ⊗ M)=aStr(M) for a ∈ A and M ∈ End(M). • The extension of the supertrace vanishes on supercommtators in A ⊗ End(M), since [a ⊗ M,b ⊗ N]=(−1)|M|·|b|ab ⊗ [M,N] wnere A is supercommutative. • If E1 and E2 are two supertraces, we may identify End(E1 ⊗E2) with End(E1)⊗ End(E2), the action being as follows: |a2|·|e1| (a1 ⊗ a2)(e1 ⊗ e2)=(−1) (a1e1) ⊗ (a2e2), where ai ∈ End(Ei) and ei ∈ Ei. ± Definition. If E is a supertrace such that dim (E )=m∓, we define its deter- minant line to be the one-dimensional vector space det(E)=(Λm+ E+)∗ ⊗ Λm− E−. • If E+ = E−, then det(E) is canonically isomorphic to R. Definition. If E = E+ ⊕E− is a superbundle, then its determinant line bundle det(E) is the bundle det(E)=(Λm+ E+)∗ ⊗ Λm− E−, ± where m± =rk(E ). • If M is an n-dimensional manifold, det (TM) is the volume-form bundle ΛnT ∗M. Definition. If E is a Hermitian supertrace, we say that u ∈ End−(E) is odd self-adjoint if it has the form ou− u = , u+ 0 where u+ : E+ → E−, and u− is the adjoint of u+. j • Let V be a real vector space with basis ei and dual basis e . Denote k k ιk = ι(ek),ε= ε(e ), k where ε is the exterior product and ιk is the interior product. k ∗ k ∗ I I – We will identify Λ V with (Λ V ) by setting he ,eJ i = δJ , where 1 k I = {1 ≤ i < ···<i ≤ dim V }, and J = {1 ≤ j1 < ···<jk ≤ dim V } I i1 ik are multiindices and e = e ∧···∧e , eJ = ej1 ∧···∧ejk . – Using this identification, it is easy to see that ε(α)∗ = ι(α) ∈ End(ΛV ∗) and that ι(ν)∗ = ε(ν) ∈ End(ΛV ). • If A ∈ End(V ), we denote by λ(A) the unique derivation of the superalgebra ΛV which coincides with A on V ⊂ ΛV ; it is given by the explicit formula j k λ(A)=Xhe ,Aekiεj ι . j,k 4 Definition. A non-zero linear map T :ΛV → R which vanishes on ΛkV for k<dim (V ) is called a Berezin integral. • Let us explain why such a linear map is called an integral. • If V is a real vector space, the superalgebra ΛV ∗ is the algebra of polynomial functions on the purely odd supertrace with E+ =0,E− = V. i i ∗ If ei is a basis for V with dual basis e , the element e ∈ ΛV play the role of coordinate function on V . – From this point of view, the interior multoplication ι(ν):ΛV ∗ → ΛV ∗,(ν ∈ V ), is the operation of differention in the direction ν. – We see that a Berezin integral is an analogue of the Lebesgue integral: a Berezin integral T :ΛV ∗ → R is a linear form on the function space ΛV ∗ which vanishes on “partial derivatives”: T · ι(ν)α =0, ∀ν ∈ V and α ∈ ΛV ∗. • If V is an oriented Euclidean vector space, there is a canonical Berezin integral, defined by projecting α ∈ ΛV onto the component of the monomial e1 ∧···∧en; here n is the dimension of V and ei form an oiented orthonormal basis of V . – We will denote the Berezin integral by T : 1, |f| = n, T (eI )= 0, otherwise. If α ∈ ΛV , we will often denote T (α)byα[n], n although strictly speaking α[n] is an element of Λ V and not of R. 2 • If A ∈ Λ V , its exponential in the algebra ΛV will be denoted by expΛ A = P(Ak/k!). Definition. The Pfaffian of an element A ∈ Λ2V is the number PfΛ(A)=T (expΛ A). The Pfaffian of an element A ∈ Lie(so(V )) is the number Pf(A)=T (expΛ XhAei,ejiei ∧ ej ). i<j 2 Example. If V = R with orthonormal basis {e1,e2}, and if 0 −θ A = θ 0 then the Phaffian of A is θ. • The Phaffian vanishes if the dimension of V is odd. • If the dimension of V is odd, Pf(A) is a polynomial of homogeneous order n/2 in the components of A. If the orientation of V is reversed, it changes sign. 5 Proposition 2. The Pfaffian f an antisymmetric linear map is a square root of the determinant (Pf(A)2)=detA. Proof. By the spectral theorem, we can choose an oriented basis ej of V such that ≤ ≤ n there are real numbers cj ,1 j 2 , for which Ae2j−1 =cj e2j Ae2j = − cj e2j−1. In this way we reduce the proof to the case in which V is the vector space R2, and Ae1 = θe2, Ae2 = −θe1. 0 −θ In this case Pf(A)=θ, while the determinant of A = is θ2. θ 0 6 Superconnections • If M is a manifold and E = E+ ⊕E− is a superbundle on M, let A(M,E) be the space of E-valued differential forms on M. – This space has a Z-grading given by the degree of a differential form, and we will denote the component of exterior degree i of α ∈A(M,E)byα[i]. – In addition, we have the total Z2-grading, which we will denote by A(M,E)=A+(M,E) ⊕A−(M,E); and which is defined by A(M,E)=A2i(M,E±) ⊕A2i+1(M,E∓); for example, the space of sections of E± is contained in A±(M,E). • If g is a bundle of Lie subalgebras on M, then A(M,g) is a Lie superalgebra with respect to the Lie superbracket defined by |X1|·|α2| [α1 ⊗ X1,α2 ⊗ X2]=(−1) (α1 ∧ α2) ⊗ [X1,X2]. Likewise, if E is a superbundle of modulus for g with respect to the action ρ, then A(M,E) is a supermodule for A(M,g), with respect to the action ρ(α ⊗ X)(β ⊗ ν)=(−1)|X|·|β|(α ∧ β) ⊗ (ρ(X)ν).
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