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Let $M$ be a differentiable manifold.

Is there a name for the maximum number of globally defined independent vector fields on $TM$ which are tangent to the fibers of $TM\to M$? Is there a name for the maximum number of globally defined independent vector fields on $TM$ which are tangent to the fibers of $TM\to M$ and whose mutual flows commute, i.e. they are vertical and have pairwise zero Lie bracket? What kind of characteristic classes can be used to compute such quantities? What are these maximum numbers for $M = S^n$?

Edit:(After the answer by Michael Albanese)

The vertical rank of $TM$ is the maximum number of independent commuting vertical vector fields on $TM$. the rank of $M$ is the maximum number of independent commuting vector fields on $M$?This terminology coined by Milnor

Question: Is the vertical rank of $TM$ equal to the rank of $M$?

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    $\begingroup$ A vector bundle has $k$ independent vector fields iff its classifying map lifts from $BO(n)$ to $BO(n-k)$. In your case, this is the composition of the classifying map of $TM$ with the projection to $M$, which is a homotopy equivalence. For spheres, this is known explicitly, cf en.m.wikipedia.org/wiki/Vector_fields_on_spheres Integrable distributions can probably also be described explicitly using Haefliger's theorem. $\endgroup$
    – Bertram Arnold
    Commented Dec 19, 2020 at 14:06

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I will address the first version of your question (i.e. no conditions on commuting flows).

A vector bundle $E \to B$ admits $k$ linearly independent vector fields if and only if $E$ has a subbundle isomorphic to $\varepsilon^k$, the trivial rank $k$ bundle. The largest such $k$ is called the span of $E$. If $E$ has rank $n$ with span $k$, then $w_i(E) = 0$ for $i > n - k$ and $p_i(E) = 0$ for $i > \lfloor\frac{1}{2}(n-k)\rfloor$.

If $\pi : TM \to M$ denotes the natural projection, then the subbundle of $TTM$ consisting of vectors tangent to the fibers of $\pi$ is precisely $\ker(d\pi)$. Your first question can be rephrased as: what is the span of $\ker(d\pi)$? Note that $\ker(d\pi) \cong \pi^*TM$, and hence $\operatorname{span}(\ker(d\pi)) = \operatorname{span}(\pi^*TM)$. In general, $\operatorname{span}(f^*E) \geq \operatorname{span}(E)$ but in our case, $\pi : TM \to M$ is a homotopy equivalence, so we obtain $\operatorname{span}(\pi^*TM) = \operatorname{span}(TM)$ which we often call the span of $M$.

The span of a manifold is very difficult to calculate in general. For spheres, the problem was resolved by Adams in 1962 - note, the characteristic class conditions mentioned above tell us nothing here as spheres are stably parallelisable. The span of $S^{n-1}$ is $\rho(n) - 1$ where $\rho(n)$ denotes the $n^{\text{th}}$ Radon-Hurwitz number, defined as follows: if $n = 2^{4a + b}c$ where $a, b, c$ are non-negative integers, $0 \leq b \leq 3$ and $c$ is odd, then $\rho(n) = 8a + 2^b$.

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  • $\begingroup$ Thank you for your very helpful answer. To be honest i was aware of isomorphicity of a vertical space at $V_p$ with $T_p M$ but I did not pay attention to the equivalent global formulation you mentioned: The subbundle of vertical tangent vectors is isomorphic to the pull back bundle $\pi^* TM$. So by your very argument we realize that if we have $k$ commuting independent vector fields on $M$ then we have $k$ commuting independent vertical vector field on $TM$. So the vertical rank of TM is $\geq$ than the rank of $M$., $\endgroup$
    – Ali Taghavi
    Commented Dec 19, 2020 at 16:42
  • $\begingroup$ By rank of $M$, I mean the maximum number of commuting independent vector fields. a terminology coined by, i think Milnor. So can one say that the vertical rank of TM is always equal to rank of M? $\endgroup$
    – Ali Taghavi
    Commented Dec 19, 2020 at 16:44
  • $\begingroup$ A more explanation of my previuos 2 comments. let $V_p$ is given. We define a linear isomorphism from $T_p M$ to vertical vectors in $T_{V_p} TM$ with $X\mapsto d/dt V_p+tX\mid_{t=0}$. But this map preserves the Lie bracket, so commuting vector fields would be lifted tocommuting vector fields. Am I right? $\endgroup$
    – Ali Taghavi
    Commented Dec 19, 2020 at 16:57
  • $\begingroup$ But I guess you some thing is missing in your notations when you write:$d\pi:\ker d\pi \to TM$ is an isomorphism", right? $\endgroup$
    – Ali Taghavi
    Commented Dec 19, 2020 at 17:07
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    $\begingroup$ @AliTaghavi: The lifts aren't canonical, but for any such lifts we have $d\pi[\tilde{X}_i, \tilde{X}_j] = [d\pi(\tilde{X}_i), d\pi(\tilde{X}_j)] = [X_i, X_j] = 0$ so $[\tilde{X}_i, \tilde{X}_j]$ is a vertical vector field, but it's not clear to me that you can choose the lifts so that the Lie bracket is zero. $\endgroup$
    – Michael Albanese
    Commented Dec 21, 2020 at 4:12

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