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**Example text**

For example when j = 1 then c1 (F ⊗ G) = rc1 (G) + sc1 (F ). To see this we apply the splitting principle and find a space Y and a map φ : Y → X such that the map H ∗ (X, Z) → H ∗ (Y, Z) is injective and both φ∗ (F ) and φ∗ (G) split into sums of line bundles: φ∗ (F ) = L1 ⊕ · · · ⊕ Lr , φ∗ (G) = L1 ⊕ · · · ⊕ Ls . , c1 (Ls )) , and φ∗ (F ⊗ G) = ⊕k,l Lk ⊗ Ll . So we obtain φ∗ cj (F ⊗ G) = σj (c1 (Lk ) + c1 (Ll ) - the polynomial in rs variables. It is well-known that σj (c1 (Lk ) + c1 (Ll )) can be expressed in terms of σi (c1 (Lk )) and σi (c1 (Ll )) .

Now we notice that when the curvature R vanishes then ∇ ∇ ∇ · · · → Γ(∧p T ∗ X ⊗ E) → Γ(∧p+1 T ∗ X ⊗ E) → · · · is a complex. Now we want to get ordinary differential forms from an End(E)valued 2 - form R. We pick a polynomial function P : Mr (C) → C 60 CHAPTER 2. COHOMOLOGY OF VECTOR BUNDLES which is homogeneous of degree j and conjugation invariant: P (B −1 CB) = P (C). When we apply P to R ∈ Γ(∧2 T ∗ X ⊗End(E)) we get a complex valued 2j-form P (R). , λr ) is the set of eigenvalues of A. This is exactly the coefficient of tj in the characteristic polynomial det(I + tA), where I is the identity matrix.

Then we get a connection ∇E + ∇F on the direct sum E ⊕ F such that the corresponding curvature matrix of 2-forms is given by R∇E +∇F = R∇E 0 0 R∇F . It is a simple fact which follows from the identity det(I + t A 0 )= 0 B det(I + tA) det(I + tB) that Pj A 0 0 B j Pk (A)Pj−k (B). = k=0 This implies that Ωj (E + F ) = jk=0 Ωk (E) ∧ Ωj−k (F ). √ (4) For a line bundle L we know that if Ω1 = (−2π −1)−1 R then its cohomology class [Ω1 ] = c1 (L) ∈ H 2 (X, C). Now to see that for each j one has [Ωj ] = cj (E) one can easily see that the classes [Ωj ] are compatible with pull-backs and then use splitting principle or the uniqueness theorem for the Chern classes.