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........Of course, this in itself does not tell us why the Clifford bundle has no continuous cross-sections. To understand this it will be helpful to look at the Clifford bundle in another way. In fact, it turns out that each point of our sphere $S^{3}$ can be interpreted as a unit-length ‘spinorial’ tangent vector to $S^{2}$ at one of its points. Recall from §11.3 that a spinorial object is quantity which, when completely rotated through 2$\pi$, becomes the negative of what it was originally. According to the above statement, a cross-section of our bundle $\text{B }\S^{3}$ would represent a continuous field of suchs pinorial unit vectors on $\text{M} S^{2}$. Now, it is a well-known topological fact that there is no global continuous field of ordinary unit tangent vectors on $S^{2}$. (This is the problem of combing the hair of a ‘spherical dog’! It is impossible for the hairs to lie flat in a continuous way, all over the sphere.) Making these directions ‘spinorial’ clearly does not help, so no global continuous field of unit spinorial tangent vectors can exist either. Hence our bundle B $S^{3} has no cross-sections.....

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Here $\text{M}$ is Manifold.

How did he conclude that Bundle B has no cross-Section?

P.S: I am new to differential Geometry.