Homology cross product and orientation

[quasar987]

An orientation of a real n dimensional vector space V is an equivalence class of an ordered basis. Equivalently, this corresponds to a choice of a generator of H_n(V,V-0)=Z. The correspondence between the two is this: given an ordered basis (v_1,...,v_n) of V, the convex hull of {v_0:=0,v_1,...,v_n} defines an n-simplex in V. Translate this n-simplex a little so that its boundary belongs to V-0 and view this as a singular n-cycle in (V,V-0). It is a generator of H_n(V,V-0). Permute two of the basis elements and you get the other generator. Reciprocally, a generator of H_n(V,V-0) is represented by a singular n-simplex as above together with a canonical order of its vertices (v_0,v_1,...,v_n). The associated orientation is the one determined by (v_1-v_0, v_2-v_0,...,v_n-v_0).

Now my question: Say µ in H_n(R^n, R^n-0) is the generator corresponding to the canonical orientation [e_1,...,e_n] of R^n and µ' in H_n'(R^n', R^n'-0) is the generator corresponding to the canonical orientation [e_1,...,e_n'] of R^n'. How to see (or prove) that µ x µ' in H_{n+n'}(R^{n+n'},R^{n+n'}-0) is the generator corresponding to the canonical orientation [e_1,...,e_{n+n'}] of R^{n+n'} ??

Thx

 

[Evalesco]

To answer your question, you need to understand the concept of direct sum. The direct sum of two vector spaces V and W, denoted V ⊕ W, is the set of all vectors (v, w) such that v ∈ V and w ∈ W. The direct sum of two generators, µ and µ', of H_n(V,V-0) and H_n'(W,W-0), respectively, is then a generator of H_{n+n'}(V ⊕ W,V ⊕ W-0). This generator is generated by the ordered basis [e_1,...,e_n, e_1',...,e_n'] which corresponds to the canonical orientation [e_1,...,e_{n+n'}] of R^{n+n'}. Thus, we can conclude that µ x µ' in H_{n+n'}(R^{n+n'},R^{n+n'}-0) is the generator corresponding to the canonical orientation [e_1,...,e_{n+n'}] of R^{n+n'}.