Let $M$ be a complex manifold and $S \subset M$ a compact complex submanifold together with a holomorphic retraction $$r : M \to S,$$ i.e. a holomorphic map which restricts to the identity on $S$.

Question. Is there a holomorphic tubular neighbourhood of $S$ in $M$?

(That is, a neighbourhood $U$ of the normal bundle of $S$ in $M$ together with a biholomorphism from $U$ to a neighbourhood of $S$ in $M$ which restricts to the identity on $S$.)


(1) If a tubular neighbourhood exists, then the bundle map gives such a retraction (after replacing $M$ by a neighbourhood of $S$ in $M$).

(2) There are well-known obstructions to holomorphic tubular neighbourhoods. For instance [1], if a tubular neighbourhood exists then the short exact sequences $$0 \to \mathcal{I}_S / \mathcal{I}_S^{k + 1} \to \mathcal{O}_M/\mathcal{I}_S^{k+1} \to \mathcal{O}_S \to 0$$ split for all $k \ge 1$. But, of course, in our case all these sequences do split via $r^* : \mathcal{O}_S \to \mathcal{O}_M$. There is a further condition in [1] called $k$-comfortably embedded, but I'm not sure how it relates to a retraction.


[1] Abate, M.; Bracci, F.; Tovena, F. Embeddings of submanifolds and normal bundles. Adv. Math. 220 (2009), no. 2, 620–656.


2 Answers 2


Question. Is there a holomorphic tubular neighbourhood of S in M?

I have a counterexample. Let $\phi:\; Z^2 \to Aut(C^n)$ be a group homomorphism, with $Aut(B)$ denote the group of holomorphic automorphisms of $C^n$. We assume that the image of $\phi$ preserves 0 and acts trivially on $T_0 C^n$. To construct $\phi$ which has infinite image take, for example, $\phi(1,0)(x, y) = (x, y + x^2)$ on $C^2$.

The quotient $M:= C \times C^n/Z^2$, with $Z^2$ acting by translations on $C$ and as $\phi$ on $C^n$ is a manifold admitting a retraction to the elliptic curve $E= C \times\{0\}/Z^2$, Since the normal bundle of $E$ is trivial, existence of "tubular neighbourhood" would imply that $M = E \times B$ locally around $E$. This would imply that there exists a holomorphic map $p:\; C \to C^n$ such that for any $a\in Z^2\subset C$ one has $p(z+a) = \phi(a)(p(z))$. For the example above, this would imply that $p$ has polynomial growth, hence it is algebraic, which is impossible.

This example is based on an idea of Loray, Thom, Touzet, who proved that for any elliptic curve with trivial tangent bundle in a formal scheme there is a canonical foliation for which this curve is a closed fiber.

Frank Loray (IRMAR), Olivier Thom (IRMAR), Frédéric Touzet (IRMAR) Two dimensional neighborhoods of elliptic curves: formal classification and foliations. https://arxiv.org/abs/1704.05214 (Theorem 3)

I have constructed a foliation with a unique closed fiber, and checked that the curve has no deformations. I guess I could have just referred to this paper instead.


In fact the short exact sequence above $\textit{doesn't}$ splits even in the case of retraction and $k=1$. What I mean is the following.

Indeed, it does split as a sequence of $\mathcal{O}_S$-modules. Now take $S$ to be the diagonal in $M=S\times S$. Put $k=1$ and consider the sequence of corresponding $\mathcal{O}_M$-modules. This ext is so called Atiyah's class. I claim that a tubular neighborhood of $S$ defines a holomorphic connection on $T_S$.

In our case the normal bundle to $S$ is the tangent bundle $T_S$. The tubular neighborhood for each $x\in S$ defines identification $f_x$ of a neighborhood $0\in T_x$ with a neighborhood of $x\in S$. Hence for each nearby $x,y\in S$ there is well-defined ${f_x}^{-1}(y)\in T_x$. Using affine parallel transport $P_{x\to y}:T_x\to T_x$ at $T_x$ where $P_{x\to y}v=v+f^{-1}(y)$, you get a natural identification of $T_x$ with $T_y$ via $df|_{f^{-1}(x)}\circ P_{x\to y}$.

Our Atiyah's class $A\in Ext^1(\mathcal{O}_S,\mathcal{I}_S/\mathcal{I}_S^2)$ is the universal one: for a sheaf $E$ over $S$ you can consider $A_E={\pi_2}_*(\pi_1^*E\otimes A)\in Ext^1_S(E,\Omega^1_S\otimes E)$, where $\pi_i$ are projections $S\times S\to S$. It is well-known that $A_E$ is complete obstruction for existence of a holomorphic connection over $E$. This happens rarely, since Atiyah class $A_{E}$ can be used for expression of Chern classes of $E$ (see Markaryan's paper on the subject).

In particular, if Chern classes of $T_S$ are non zero then there is no holomorphic connection and, as a corollary, there is no tubular neighborhood.

PS From this point the existence of a tubular neighborhood of the diagonal may imply vanishing of a very rich structure provided by derived co-Lie algebra defined by $A_{\Omega}$. It would be nice to see if there are examples of proper diagonals admitting a tubular neighborhood other than abelian varieties.

  • 1
    $\begingroup$ Thanks for your answer. I do not understand the first step of the definition of the connection, namely: "The tubular neighborhood for each $x\in S$ defines identification $f_x$ of a neighborhood $0\in T_x$ with a neighborhood of $x\in S$". The tubular neighbourhood gives an embedding of a neighborhood of $0$ in $T_xS$ in $S \times S$, so presumably you want to compose this with one of the projections to get the desired identification. But it is easy to find examples where this composition is not an isomorphism. $\endgroup$ Oct 6, 2021 at 17:26
  • $\begingroup$ @SimonParker, oops, you absolutely right. I unintentionally took the retraction into account and impose stronger conditions on the definition of a tubular neighborhood. Namely, if the tubular neighborhood morphism restricted to $S$ maps tangent vectors along the normal bundle to tangent vectors along the retraction fibers, then the projection you mentioned above gives an isomorphism and we can obtain a connection. $\endgroup$ Oct 8, 2021 at 12:08
  • $\begingroup$ Let $\phi:Tot(\nu_{S/M})\to Tube(S/M)$ be any tubular neighborhood. Consider a "retraction" fibers $F_x=\phi^{-1}(x)$ for $x\in S$. Tangent vectors along $F_x$ define a vector bundle $\tau_r\subset {T_M}|_{S}=\nu_{S/M}\oplus T_S$, which is transversal to $T_S$. In other words $\tau_r$ is a deformation of $\nu_{S/M}$ defined by an element in $r\in Hom(\nu_{S/M},T_S)$. If one can produce an automorphism $\alpha$ of $Tot(\nu_{S/M})$ which differential maps $\tau_r$ to $\nu_{S/M}$, then we are done. I don't see how one can do this. $\endgroup$ Oct 8, 2021 at 12:09
  • $\begingroup$ In order to make my answer not completely useless let me mention a counter-example to your original question, where the projection definitely gives an isomoprhism. Take $S=\mathbb{P}^1$. Then tangent vectors to $F_x$ ($\tau_r$) are determined by some $r\in Hom(\nu_{S/M},T_S)=\mathbb{C}$. Considering a diagonal action of $SL(2)$ on $M$ we obtain an action on $\mathbb{C}$ which is trivial, hence $\tau_r$ is invariant under the action and at least one of two projections is isomorphism. $\endgroup$ Oct 8, 2021 at 12:13

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