3
$\begingroup$

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.

Improve this question
$\endgroup$
9
  • $\begingroup$ There is a construction of a degree $k$ map from $S^n \to S^n$ in example 2.31 of Hatcher's book here math.cornell.edu/~hatcher/AT/ATch2.pdf $\endgroup$
    – PVAL
    Commented Aug 7, 2015 at 18:13
  • $\begingroup$ @PVAL Thanks, but I am looking for a construction which would be specifically for degrees of the form I mentioned, i.e., $d(d-1)/2$. I think that, as there is for $d^2$, there might be some particularly nice construction for such degrees. $\endgroup$
    – Hugh Thomas
    Commented Aug 7, 2015 at 18:24
  • 1
    $\begingroup$ A short answer is to take the join of maps $S^1 \to S^1$ of degree $d$ and $\tfrac{d-1}{2}$ if $d$ is odd, and $\tfrac{d}{2}$ and $d-1$ if $d$ is even. The join is an isomorphism $S^{n-1} * S^{m-1} \to S^{n+m-1}$, and it's more closely related to the operation $(V,W) \mapsto V \times W$ on vector spaces than the tensor product: the join connects the unit sphere in $V$ to that in $W$. However, these are probably not getting at your underlying question. $\endgroup$
    – Tyler Lawson
    Commented Aug 7, 2015 at 18:32
  • $\begingroup$ @TylerLawson Thanks. I guess I should have said that I am thinking of $d(d-1)/2$ as a binomial coefficient, and I would like a construction that in some way reflects that. $\endgroup$
    – Hugh Thomas
    Commented Aug 7, 2015 at 18:38
  • 1
    $\begingroup$ @FernandoMuro: The construction that I gave for a cycle of degree $d^2$ can be generalized to give a cycle of degree $d^n$ in $S^{2n-1}$. Another feature which I would like is that it have should generalize in a similar way, giving a class of degree $d \choose n$ in $H_{2n-1}(S^{2n-1})$. $\endgroup$
    – Hugh Thomas
    Commented Aug 8, 2015 at 3:14

1 Answer 1

Reset to default
1
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .