Let V be a 2-dimensional (real or complex) vector space with 2 generators {C,S}. The trigonometric coalgebra structure on V is defined by the commultiplication $\Delta(C)=C\otimes C-S\otimes S$ and $\Delta(S)=S\otimes C+ C\otimes S$. In this formulation, C and S play the role of standard cos and sin, repectively . Moreover .the operator of "differentialtion", which is denoted by "T" , is a nonscalar operator which satisfies
(T\otimes T)\circ \Delta=\Delta \circ T^2.
So R^2 or C^2 can be equiped with this structure where C and S are identified with (1,0) and (0,1), respectively.
I have two questions, the first question is a real question and the second is a complex one. Finally I unify these two questions in a thirth question in order to find a poissible method of differentiation of sections of certain real or complex bundle.
1)Can the structure group of the tangent bundle of a (non parallelizable) compact surface be reduced to the following subgroup of GL(2,R): Those operatores which preserves the coalgebra structure of R^2?
2)Can the structure group of the canonical 2-plane bundle of complex grassmanian G(2, n) be reduced to subgroup of GL(2,C) which preserve the coalgebra structure of C^2?
If the answer to these question is affirmative, then fibres of our bundles are "coalgebra". So our next question is the following:
3)Is there a nonscalar morphism $T$ on each of the above 2 dimensional bundles(real or complex) which satisfies
$(T\otimes T)\circ \Delta=\Delta \circ T^2$?
For any such nonscalar operators $T$, one can consider a linear operator on the space of sections of the vector bundle with
$ S \mapsto T\circ S $
This operator can be counted as a way of "differentiation" of sections.
Is there any reference which considered coalgebra structures on vector bundles?
1) If the surface is orrientable, yes. Just note that C and S also have the interpretation of Re and Im. So what you are asking is, is there a complex structure J: TM -> TM, with J^2 = -1 on the surface M . Then a pair C, S would just be local section X, JX of the tangent bundle. This is true because every orientable surface can be given the structure of a complex surface, i.e. GL(2,R) can be reduced to C^*, in fact a complex structure on a Riemann surface is the same thing as a conformal metric. Note that you may also think of the real tangentbundle as the real co tangentbundle, and then identify whis with the complex co tangent bundle often written as K. This will be important below,
2) No, by the above we are seeking a complex linear operator J on the universal subbundle S of rank 2 with J^2 = -1. Over the complex numbers such a bundle would immediately split up S in a +i and -i eigenvalue. i.e a direct sum of linebundles. But S does not split. This follows for example from the Chern classes:
c(S) = 1 + x_1 + x_2
c(Q) = 1 + y_1 + y_2 + .... y_{n-2}
H^*(Gr(2, n)) = Z[x_1, y_1, x_2, y_2, ....y_{n-2} ]/(c(S)c(Q) = 1)
i.e. y_1 = -x_1, y_2 = -x_2 + x_1^2 , y_3 = 2x_1x_2 - x_1^3, ..
now if S = L_1 + L_2 then c_1(S) = c_1(L_1) + c_1(L_2) = x_1
so c_1(L_1) = ax_1 , c_1(L_2) = (1 - a)x_1
but c_2(S) = x_2 = c_1(L_1)c_1(L_2) = a(1-a)x_1^2
which is only possible if x_1^2 is a multiple of x_2 (i.e n =3, Gr(2,C^3) = CP^2) and then we must have a(1-a) = 1 which is impossible for a in Z
3) have to think about this.
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