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I am trying to visualize a seven sphere here: http://www.youtube.com/watch?v=II-maE5HEj0. What is $(u,v)\to(u, u^hvu^j)$ in the attached picture?

Are quaternion maps commonly used to glue boundaries of topological spaces? How do we know that this quaternion is gluing the boundaries? Is $M$ a 7-sphere?

Any answers/insights would be helpful.

The Substitute
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In the picture, $M$ is being described as the total space of an $S^3$-bundle over $S^4$ with structure group $\rm{SO}(4)$, which is denoted by $\xi_{h,j}$. The bundle $\xi_{h,j}$ is trivialized over two hemispheres of $S^4$, each of which are diffeomorphic to the disk $D^4$. Hence to completely determine $\xi_{h,j}$, we just need to say what the transition function $\theta: \partial D^4 \longrightarrow \rm{SO}(4)$ on the equator is. Here the equator is $S^3$, which is identified as the group of unit quaternions. The transition function is then given by $$\theta(u)v = u^h v u^j,$$ where the multiplication is quaternion multiplication. Since unit quaternions form a group under quaternion multiplication, $\theta(u)$ maps $S^3$ to $S^3$.

The fact that the total space $M$ is a $7$-sphere (when $h + j = 1$) follows after constructing a Morse function on $M$ that has only two critical points (any closed, smooth manifold admitting a Morse function with only two critical points is homeomorphic to a sphere).

The construction in your picture is featured heavily in the classic paper On manifolds homeomorphic to the $7$-sphere by Milnor.

Henry T. Horton
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  • I believe $M$ is only homeomorphic to $S^7$ when $h+j = \pm 1$. Today, one could also quote the Poincare conjecture to learn when $M$ is homeomorphic to $S^7$ because showing it's homotopy equivalent to $S^7$ (when $h+j = \pm 1$) is easy. – Jason DeVito - on hiatus Mar 17 '13 at 01:17
  • @Henry, What does this multiplication mean in this context? – The Substitute Mar 18 '13 at 23:41
  • A quaternion is of the form $a + bi + cj + dk$, where $i^2 = j^2 = k^2 = -1$, $ijk = -1$, and $a, b, c, d \in \Bbb R$. Note that $i$, $j$, and $k$ pairwise anticommute, and that $ij = k$, $jk = i$, and $ki = j$ (these relations can be derived from the first sentence of this comment). From here we get that the product of two quaternions is\begin{align} (a_1+b_1i+c_1j+d_1k)(a_2+b_2i+c_2j+d_2k) = & a_1a_2 - b_1b_2 - c_1c_2 - d_1d_2 \ & +(a_1b_2 + b_1a_2 + c_1d_2 - d_1c_2)i\ & +(a_1c_2 - b_1d_2 + c_1a_2 + d_1b_2)j\ & +(a_1d_2 + b_1c_2 - c_1b_2 + d_1a_2)k. \end{align} – Henry T. Horton Mar 18 '13 at 23:52
  • So the product here considers $S^3, D^4 \subset \Bbb R^4$ and identifies $(a,b,c,d) \in \Bbb R^4$ with the quaternion $a + bi + cj + dk$. The product is the quaternion multiplication in my previous comment. – Henry T. Horton Mar 19 '13 at 00:00