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Introduction to Frobenius Manifolds INTRODUCTION TO FROBENIUS MANIFOLDS NOTES FOR THE MRI MASTER CLASS 2009 1. FROBENIUS ALGEBRAS Definition and examples. For the moment we fix a field k which contains Q, such as R, C or Q itself. To us a k-algebra is simply a k-vector space A which comes with k-bilinear map (the product) A × A A, (a; b) 7 ab which is associative ((ab)c = a(bc) for all a; b; c 2 A) and a unit element e 2 A for that product: e:a = e:1 = e for all a 2 A. We use! e to embed!k in A by λ 2 k 7 λe and this is why we often write 1 instead of e. Definition 1.1. Let A be a k-algebra that is commutative, associative and fi- nite dimensional! as a k-vector space. A trace map on A is a k-linear function I : A k with the property that the map (a; b) 2 A × A 7 g(a; b) := I(ab) is nondegenerate as a bilinear form. In other words, the resulting map ∗ a 7 !I(a:-) is a k-linear isomorphism of A onto the space! A of k-linear forms on A. The pair (A; I) is the called a Frobenius k-algebra. Remark! 1.2. This terminology can be a bit confusing: for a finite dimen- sional k-algebra A, one defines its trace Tr : A k as the map which assigns to a 2 A the trace of the operator x 2 A 7 ax 2 A. This is in general not a trace map. ! Exercise 1. Prove that g then satisfies g(ab; c) = g!(a; bc) and that con- versely, any nondegenerate k-bilinear symmetric map g : A × A k with that property determines a trace map on A. Exercise 2. Let (A; I) be a Frobenius algebra. Show that the product! deter- mines the unit element: no other element than e can serve as a unit. Examples 1.3. (a) A trace map on the field k (viewed as k-algebra) is given by a nonzero scalar λ, for it will be of the form a 2 k 7 λa 2 k. It is simple, meaning that is has no nontrivial quotients. n (b) Let A = k[t]=(t ) with n a positive integer. A k!-linear form I : A k is a trace map if and only if I(tn-1) 6= 0. (c) Let M be a compact oriented manifold of even dimension 2d !(for instance a compact complex manifold). Then its even degree cohomology even d 2k H (M; k) = ⊕k=0H (M; k) is a Frobenius R-algebra for the cup product and for integration: I is zero in degree 6= 2d and sends a class in degree 2d to its value on the fundamental class (if k = R, and if we use De Rham cohomology, then this is just integration of a 2d-form); the nondegenaracy 1 2 NOTES FOR THE MRI MASTER CLASS 2009 of (a; b) 7 I(a; b) simply expresses the Poincar´e duality: the fact that (a; b) 2 H2k(M; k) × H2d-2k(M; k) 7 (a [ b)[M] 2 k ! d is perfect pairing. For example, for complex projective d-space, P , d ≥ 1, d • d Heven(P ; k) = H (P ; k) is as an algebra generated! by the hyperplane class 2 d d+1 d d η 2 H (P ; k). Clearly η . On the other hand, η [P ] = 1 (this expresses d the fact that d general hyperplanes in P have exactly one point in common) and so this reproduces in fact a special case of example (b) with n = d + 1. (d) (Zero-dimensional complete intersections) Here is an interesting ex- ample from commutative algebra that generalizes example (b) above. Let f; : : : ; fm 2 k[[z1; : : : ; zm]] be formal power series without constant term and assume that they span an ideal in k[[z1; : : : ; zm]] of finite codimension: A := k[[z1; : : : ; zm]]=(f1; : : : ; fm) is finite dimensional. Then one can show that the jacobian @f det i @zj i:j has a nonzero image in A and generates there a one-dimensional ideal. A linear form I : A k gives A a trace map if and only if it is nonzero on this image. (Grothendieck defined a specific trace map on A as an iterated residue; it is written! gdf1 ^ ··· ^ dfm I(g) = Res(fm=0) ··· Res(f1=0) :) f1 ··· fm n n-1 Exercise 3. Let f := t + a1t + ··· + a0 2 k[t] be a monic polynomial and put A = k[t]=(f). The images of 1; t; : : : ; tn-1 form a k-basis of A. Prove that a trace map is given by the function I : A k that takes the value 1 on the image of tn-1 and zero on the images of lower powers of t. We next discuss three ways of producing new! Frobenius algebras out of old: direct sums, tensor products and rescalings. Direct sums. Let A and B be commutative k-algebras. Then the vector space A⊕B is an algebra for componentwise multiplication: (a; b)(a0; b0) = (aa0; bb0) with (1; 1) as its identity element. The identity element of A corresponds to ι = (1; 0), which is in A ⊕ B an idempotent: its satisfies ιι = ι. We can recognize that an algebra is thus obtained by its idempotents: if we are given a commutative k-algebra C with unit and a nonzero idempotent ι 2 C, then C0 := ιC is multiplicatively closed and ι serves as a unit element for ιC. The element 1-ι is also an idempotent, for (1-ι)2 = 1-2ι+ι2 = 1-ι, and so the same can be said about C00 := (1 - ι)C. We have C = C0 ⊕ C00 (write c 2 C as ιc+(1-ι)c), not just as k-vector spaces, but even as algebras. Lemma-definition 1.4. Let A be a finite dimensional commutative k-algebra. Then A is called semisimple if the following equivalent conditions are fulfilled: (i) A is the sum of its one-dimensional ideals, (ii) A is isomorphic to kn with componentwise multiplication. INTRODUCTION TO FROBENIUS MANIFOLDS 3 In that case A is in fact the direct sum of its one-dimensional ideals. Proof. (ii) (i). The one-dimensional ideals of kn are easily seen to be the individual summands. (i) (ii).) If I and J are distinct one-dimensional ideals of A, then we have IJ ⊂ I \ J = f0g. If A is spanned by one-dimensional ideals, then we can select) I1;:::;In of those such that A is their direct sum as a k-vector space. Write 1 = a1 + ··· + an with ai 2 Ii. If ei 2 Ii is nonzero, then ei = ei:1 = j eiaj = eiai. It follows that ai 6= 0 and so ai generates Ii also. Upon replacing ei by ai, we see that ai is an idempotent and that P (λ1; : : : ; λn) 7 i λiai is an isomorphism of algebras. Exercise 4. ProveP that for a semisimple commutative k-algebra A, the trace Tr : A k (which! assigns to a the trace of x 2 A 7 ax 2 A) is also a trace map. Compute it in terms of its decomposition into one-dimensional ideals. If (A;! I) and (B; J) are Frobenius k-algebras, then!(a; b) 7 I(a) + J(b) is a trace map on A ⊕ B. It is easy to check that any trace map on the k-algebra A ⊕ B must be of this form. In particular, a trace map on! a semisimple k- algebra of dimension n is given by n nonzero scalars. More precisely, it is given by a map from the n-element set of its one-dimensional ideals to k×. Remark 1.5. In the situation of Lemma-definition ??, the maximal ideals of A are the direct sums of all but one of the one-dimensional ideals. So these are also n in number and the corresponding quotient algebras are one-dimensional. If you are familiar with the language of algebraic geome- try, then the above Lemma says that A is semisimple if and only if is it the coordinate ring of a reduced k-variety Spec(A) that has n distinct points. An idempotent of A generating one dimensional ideal is the the character- istic function of a point of Spec(A) (the maximal ideal defining the point is the sum of the remaining one dimensional ideals). So a trace map can be interpretated as a function Spec(A) k×. Tensor product. Let A and B again be commutative k-algebras. Then the vector space A⊗B is an algebra whose! product is characterized by the prop- erty that (a⊗b)(a0⊗b0) = aa0⊗bb0 (beware that a general element of A⊗B is of the form i ai ⊗ bi). Its unit is 1 ⊗ 1. If (A; I) and (B; J) are Frobenius algebras, then we have a trace map on A ⊗ B defined by (a ⊗ b) 7 I(a)J(b). This is best seenP by using the associated nondegenerate symmetric bilinear 0 0 0 0 forms g(a; a ) := I(aa ) and h(b; b ) := J(bb ). For then it reduces! to seeing that ((a; b); (a0; b0)) 7 g(a; a0)h(b; b0) factors through a symmetric bilinear form (A ⊗ B) × (A ⊗ B) k that is nondegenerate whenever g and h are. This property no longer! involves the algebra structure and is easily verified by choosing bases of A resp.! B on which g resp.
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