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In a survey paper of Thom's work, Sullivan discusses an example of an isolated singularity (see p. 5). We begin by considering rank one 2x3 matrices which can be viewed as a 4-fold in $\mathbb{C}^6$; projectivizing, we get a 3-fold $V \subset \mathbb{CP}^5$. Sullivan then says that we may consider the complex cone of $V$ inside of $\mathbb{C}^6$. My guess is that this object can be defined via $C(V):= \{(v,x,y) \in \mathbb{C}^4 \times \mathbb{C}^2:|v|^2 + x^2+y^2=0 \}$ but then we need to know that $V$ can be embedded in $\mathbb{C}^4$.

Q1: Is that the correct definition of a complex cone or am I way off base here?

We now want to obtain an element in $\pi_{11}(MU(2))$ associated to this singularity. We then take the link $L=S^{11} \cap C(V)$ since, Sullivan claims, the cone has an isolated singularity. My guess is that since the complex tangencies are well-defined, we can map $L \to \text{Gr}_\mathbb{C}(2,6)$ via $x \in L \mapsto \mathbb{C}^6/T_x C(V)$, the normal complex 2-plane. This includes into $BU(2)$ and we can extend the map to $S^{11}$ by mapping everything in the complement of $L$ to a point. Lifting this map to the tautological bundle $\xi_2 \to BU(2)$ and then into the Thom space $MU(2)$, we get an element of $\pi_{11}(MU(2))$.

Q2: This seems the most natural way to obtain an element $\pi_{11}(MU(2))$ but is it the one that Sullivan means? Or more to the point, I don't know whether this element is nontrivial over mod 2 and rationally. I don't have a copy of Hartshorne's book and it seems the Weil paper is in Italian so I am unable to follow-up with the listed references.

Q3: One conclusion is that $L$ does not bound a stably almost complex manifold in the 12-ball. Is there any way to conclude that it does not bound any SAC manifold, independent of any ambient space?

Q4: Is it known what 7-manifold $L$ is, topologically and smoothly? I'm asking just in case it's an exotic 7-sphere or something.

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    $\begingroup$ For Q3) the answer is that it does bound an SAC manifold, because the dim 7 complex bordism ring is zero $\endgroup$
    – kiran
    Dec 17, 2021 at 17:27

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