Constructing a homology class of degree $d(d-1)/2$ in $H_3(S^3)$ There is a nice construction of a class of degree $d^2$ in $H_3(S^3)$. Take a class $h$ of degree $d$ in $H_1(S^1)$, and then take its join with itself: $h*h$ is degree $d^2$ in $H_3(S^1*S^1)$, and $S^1 * S^1$ is $S^3$.  
I would like a similar construction of a class whose degree is $d(d-1)/2$.  The difference between the construction I described above and the one I want, seems formally like the distinction between tensor product and exterior product, but I don't know how to make any sense of that.  
 A: I am coming around to FernandoMuro's point of view that there are
a multitude of solutions, among which I haven't managed to specify 
sufficiently clearly which one(s) interest me.  (Thanks, FernandoMuro, and the other commenters, for your helpful responses.)   
For example, take any 3-dimensional simplicial ball with $d \choose 2$ 
facets.  The space obtained by quotienting out the boundary is an 
$S^3$, and it admits a degree $d \choose 2$ map to a $S^3$ by sending each
simplex to $S^3$, sending the boundary of each simplex 
to a single point and the interior of each simplex to cover the $S^3$ (each with the same orientation).  
I will record one specific example of such a simplical ball, because I rather like it.  Consider the simplicial complex whose vertex set is $\{1,\dots,d+2\}$, and whose facets are $\{i,i+1,j,j+1\}$ with
$1\leq i$, $\ i+1<j$, and $j+1\leq d+2$.  This is a triangulation of a 3-dimensional cyclic polytope with $d+2$ vertices; in fact it is the unique
such triangulation with $d\choose 2$ simplices.  (One sees that it is a
triangulation because it consists of the lower facets of a 4-dimensional cyclic polytope when one embeds the cyclic polytope in $\mathbb R^4$ in the usual way.)
I would still be interested in other solutions if anyone has any suggestions.
