Classification of fibrations $\Bbb S^k\longrightarrow\Bbb S^d\longrightarrow B$

Question

Does there exist a complete classification of all fiber bundles $\Bbb S^k\longrightarrow\Bbb S^d\longrightarrow B$, that is, fibrations of $\smash{\Bbb S^d}$ with each fiber homeomorphic to $\smash{\Bbb S^k}$ for some fixed $k\le d$.

The Wikipedia page on Hopf fibrations contains a list of some real/complex/quaternionic/octonionic fibrations. In other words: is this list complete?

I am then interested, which of the base spaces $B$ that appear in above classification admit a topological/Lie group structure (compatible with its present topology).

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Update

The last part of my question about topological/Lie group structure (which should have been a separate question from the start) was partially answered here. The projective spaces listed there are exactly the base spaces of sphere fibrations by great spheres (according to "On fibrations with flat fibres" by Ovsienko and Tabachnikov). It says nothing about the general case, though.

Answer 1

I'll assume that $1\leq k<d$, the other cases being easy. Then the long exact sequence of homotopy groups shows that $B$ is simply connected, so we have a Serre spectral sequence with untwisted coefficients: $$ E_2^{ij} = H^i(B)\otimes H^j(S^k) \Longrightarrow H^{i+j}(S^d), $$ with $d_r\colon E_r^{ij}\to E_r^{i+r,j-r+1}$. Let $u$ and $v$ be the generators of $H^k(S^k)$ and $H^d(S^d)$. The only possible differential is $d_{k+1}\colon H^i(B)u\to H^{i+k+1}(B)$, and this is has the form $au\mapsto ax$ for some $x\in H^{k+1}(B)$. The only way the spectral sequence can converge to $H^*(S^d)$ is if $k$ is odd and $H^*(B)=\mathbb{Z}[x]/x^{r+1}$ with $(r+1)(k+1)=v+1$. We can now apply Adams's theorem on elements of Hopf invariant one to the first attaching map in $B$ to see that $k\in\{1,3,7\}$. I think that there is an argument along similar lines that if $k=7$ we can only have $r\leq 2$, but I don't remember details. Thus, your fibration looks cohomologically like one of the standard fibrations $S^1\to S^{2r+1}\to \mathbb{C}P^r$ or $S^3\to S^{4r+3}\to \mathbb{H}P^r$ or $S^7\to S^{8r+1}\to\mathbb{O}P^r$. For the case $k=1$ we can use $[X,\mathbb{C}P^\infty]=H^2(X)$ to get a map $B\to\mathbb{C}P^\infty$ and check that it restricts to give a homotopy equivalence $B\to\mathbb{C}P^r$. In the other cases I think it is also true that $B$ is homotopy equivalent to $\mathbb{H}P^r$ or $\mathbb{O}P^r$, but the argument is more complicated and again I do not remember details. [UPDATE: See comment below from Igor Belegradek: this last sentence is apparently wrong.]