What is the generator of $\pi_9(S^3)$? I have Toda's book Composition methods in homotopy groups of sphere.
I have not found the generator of $\pi_9(S^3)=\mathbb{Z}_3$.
What is it?
 A: Toda's sequence
$$S^3 \to \Omega \widehat{S}^4 \to \Omega S^{11}$$
is a fibration $3$-locally.  I claim that $(\pi_9 \Omega\widehat{S}^4)_{(3)}=0$ in which case your element must be in the image of $\pi_{10}\Omega S^{11}=Z$ under the boundary map of the above sequence.  
To prove my claim, I appeal to (i) the $3$-primary fibration sequence $\widehat{S}^4\to \Omega S^5\to \Omega S^{13}$, and (ii) wikipedia, who tells me that there is no $3$-torsion in the $6$-stem of $S^5$ (or, just use more Toda sequences to get up to the stable range.)
A: A generator may be deduced indirectly as follows (using only information available in Hatcher's Algebraic Topology (AT)):
One has a fiber bundle $$Sp(n-1)\to Sp(n)\to S^{4n-1}$$ (p. 383, Example 4.55 of AT ). Taking $n=2$, we
have a fiber bundle $$S^3=\mathbb{H}^1=Sp(1)\to Sp(2) \to S^7.$$
From the long exact sequence of homotopy groups (Thm. 4.41 AT), we have
$$\cdots \to \pi_{10}(S^7)\to \pi_9 Sp(1)\to \pi_9 Sp(2) \to \cdots$$
which is exact at $\pi_9 Sp(1)$. 
Again from the fiber bundles 
$$Sp(2)\to Sp(3)\to S^{11}, Sp(3)\to Sp(4)\to S^{15}, \ldots$$ and long exact sequence, we see that we're in the
stable range, so $$\pi_9 Sp(2)=\pi_9 Sp(3) = \cdots = \pi_9 Sp(\infty) = \pi_5 O(\infty)= 0$$ by Bott periodicity (p. 384 Hatcher). 
So we have a surjection $$\mathbb{Z}_{24} =\pi_{10} S^7\twoheadrightarrow \pi_9 S^3.$$
By the Freudenthal suspension theorem (p. 360, Cor. 4.24 Hatcher), and from the table on p. 339 of AT, 
$$\mathbb{Z} + \mathbb{Z}_{12} \cong \pi_7 S^4 \twoheadrightarrow \pi_8 S^5 \cong \pi_9 S^6 \cong \pi_{10} S^7 \cong \mathbb{Z}_{24}.$$
Moreover, the $\mathbb{Z}$
factor of $\pi_7 S^4$ is generated by the Hopf map $\nu$ (Example 4.46, 1st paragraph p. 385 AT) associated
to the quaternions $$S^3 \to S^7 \overset{\nu}{\to} \mathbb{HP}^1=S^4$$  Hence $\langle \nu \rangle \cong \mathbb{Z} \leq \pi_7 S^4$ must surject $\mathbb{Z}_{24}=\pi_8 S^5$ under the suspension map. 
Summarizing, we see that 
$$ \pi_7 S^4 \geq  \langle \nu\rangle \twoheadrightarrow \pi_{10} S^7 \twoheadrightarrow \pi_9 S^3,$$
where the first surjection comes from triple suspension, and the
last map is the connecting map in the fibration exact sequence. 
The connecting map is induced from the isomorphism and boundary map $\pi_{10} S^7 \cong \pi_{10} (Sp(2),Sp(1)) \overset{\partial}{\to} \pi_9 Sp(1)$. To understand this geometrically, I suppose one has to understand the proof of the isomorphism, which requires a lift $D^{10} \to Sp(2)$ lifting the map $D^{10} \to S^7$. I believe that it might be possible to describe this map fairly explicitly using coordinates, eg using a connection for the bundle. 
A: Here's another way to describe the generator of $\pi_9(S^3)\cong\mathbb{Z}/3$.
By Proposition 13.6 in Toda's book, and the remarks preceding it, a generator is given by the composition
$$
S^9 \stackrel{\alpha_1(6)}{\longrightarrow} S^6 \stackrel{\alpha_1(3)}{\longrightarrow} S^3
$$
where $\alpha_1(3)\in \pi_6(S^3)\cong\mathbb{Z}_{12}$ is a generator of the $3$-primary component and $\alpha_1(6)=E^3\alpha_1(3) \in \pi_9(S^6)\cong\mathbb{Z}_{24}$ is its third suspension. 
The element $\alpha_1(3)$ has a very nice geometric representative described in my answer here. Explicitly, let $\mathbb{H}$ denote the quaternions, and represent the $6$-sphere as
$$
S^6 = \{(p,w)\in \mathbb{H}\times\mathbb{H} \mid \mathfrak{Re}(p)=0\mbox{ and } |p|^2+|w|^2=1\}.
$$
The map $b:S^6\to S^3\subseteq \mathbb{H}$ is given by 
$$
b(p,w) = \left\{\begin{array}{ll} \frac{w}{|w|} e^{\pi p} \frac{\overline w}{|w|}, & w\neq 0 \\
-1, & w=0, \end{array}\right.  
$$
where $e^{\pi p} = \cos(\pi |p|) + \sin(\pi|p|) \dfrac{p}{|p|}$ is the quaternionic exponential.  
This description appears in the paper
Abresch, U.; Durán, C.; Püttmann, T.; Rigas, A., Wiedersehen metrics and exotic involutions of Euclidean spheres, J. Reine Angew. Math. 605, 1-21 (2007). ZBL1125.57017.  
