Compile script bundle into app from CLI

I have an AppleScript (JXA) bundle with several resources inside that talk to each other. I can open it and export to an app from the GUI, but I’m looking for a way to do so from the command line. I’ve looked into osacompile, but it isn’t including the resources. All the information I see pertains to converting single scripts to apps, never script bundles.

Is there a way to do it?

What is a conic bundle and why is it called so?

I am desperately trying to understand what is a conic bundle. It seems like this is a completely standard term in algebraic geometry, there is even a page on wiki about it, but this doesn’t really help. For example, I have the following test question.

Test question. Let $ S$ be a smooth complex ruled surface $ S\to C$ over a curve $ C$ . Suppose we blow up $ S$ twice in the same fibre, so that the preimage of one point in $ C$ is a union of three lines. Is this a conic bundle or not?

My problem is the following. According some definitions that I saw, in a conic bundle the preimage of a point should be a conic. And a union of three lines is not a conic. However, I tried to trace back the definition of conic bundles, and one of the earliest versions that I found is an article of Srakisov 1980 (in Russian): http://www.mathnet.ru/links/b10a1373601dacdd9b7debba2b3e1c8f/im1862.pdf

Sarkisov is just asking that the preimage of a generic point be a rational curve.

I fear that my main question (what is a conic bundle) is a complicated one, for example judging by the fact that the answer to the following question was not given by the mathoverflow community: References about conic bundles

Question 2. Why Sarkisov calls his bunldes conic bundles? Is there some relatively pedagogical place where on can read about this?

(I fear that I can’t understand this because I misinterpret the expression “generic fiber”)

How to prove two curves in the frame bundle to project to the same curve on base manifold?

There is a problem about Cartan’s development, arising from the paper ‘Kinetic Brownian motion on Riemannian manifolds’, Subsection 2.4.1. To be precise, let $ (M,g)$ be a $ d$ -dimensional complete Riemannian manifold, $ \pi:OM\to M$ be its orthonormal frame bundle with structure group $ O(d)$ . Denote by $ H_v$ the standard horizontal vector field on $ P$ corresponding to $ v\in\mathbf R^d$ , uniquely characterized by the property that $ \pi_*(H_v(z)) = e(v)$ for all $ z = (x,e) \in OM$ . Let $ \{\epsilon_1,…,\epsilon_d\}$ be the canonical basis of $ \mathbf R^d$ , with dual basis $ \{\epsilon_1^*,…,\epsilon_d^*\}$ . Denote by $ V_i, 1\le i\le d$ the vertical vector field induced by $ a_i=\epsilon_i \otimes \epsilon_1^* – \epsilon_1 \otimes \epsilon_i^* \in SO(d)$ .

Given a smooth curve $ \{m_t\}_{0\le t\le1}$ , define the Cartan’s development of $ \gamma$ on $ OM$ as the solution to the ODE on $ OM$ , \begin{equation}\tag{1} \dot z_t = H_{\dot m_t}(z_t), \quad z_0 = (x_0,e_0)\in OM. \end{equation}

Now Assume $ \{m_t\}_{0\le t\le1}$ is run at unit speed, i.e., $ |\dot m_t|\equiv 1$ . Then, given an orthonormal basis $ f_0$ of $ \mathbf R^d$ with $ f_0(\epsilon_1) = \dot m_0$ , solve the following ODE on $ SO(d)$ , \begin{equation} \dot f_t = \sum_{i=2}^d (f_t(\epsilon_i),\ddot m_t)a_i(f_t), \end{equation} started from $ f_0$ , and define the $ \mathbf R^{d-1}$ -valued path $ \{h_t\}_{0\le t\le1}$ , starting from zero, by the ODEs \begin{equation} \dot h^i_t = (f_t(\epsilon_i),\ddot m_t), \quad 2\le i\le d. \end{equation} Consider the following ODE on $ OM$ , \begin{equation}\tag{2} \dot{\tilde z}_t = H_{\epsilon_1}(\tilde z_t)+ \sum_{i=2}^d V_i(\tilde z_t) \dot h_t^i, \quad \tilde z_0 = (x_0,e_0)\in OM. \end{equation} Then the paper, mentioned in the very beginning, has the following claim:

Claim: $ \pi(\tilde z_t) = \pi(z_t)$ .

But why?


I try to prove this claim. But I am not able to finish that.

Use the coordinate system $ (x^i,e_l^k)$ on $ OM$ . Then \begin{align} H_v &= v^j e_j^i \partial_{x^i} – v^r \Gamma^k_{ij} e_l^j e_r^i \partial_{e_l^k}, \ V_i &= e_i^k \partial_{e^k_1} – e_1^k \partial_{e_i^k}. \end{align} On the one hand, Eqn. (1) is represented as \begin{equation}\left\{ \begin{aligned} \dot{x}^i &= e_j^i \dot m^j, \ \dot e_l^k &= -\Gamma^k_{ij} e_l^j e_r^i \dot m^r, \end{aligned} \right. \end{equation} where $ \Gamma^k_{ij}$ are the Christoffel’s symbols of the metric $ g$ . We can obtain \begin{equation} \ddot x^i = \dot e_j^i \dot m^j + e_j^i \ddot m^j = -\Gamma^i_{kl} e_j^l e_r^k \dot m^r \dot m^j + e_j^i \ddot m^j = -\Gamma^i_{kl} \dot x^l \dot x^k + e_j^i \ddot m^j, \end{equation} that is, \begin{equation}\tag{1*} \frac{\nabla \dot x^i}{dt} = e_j^i \ddot m^j. \end{equation} On the other hand, Eqn. (2) is represented as \begin{equation}\left\{ \begin{aligned} \dot{\tilde x}^i &= \tilde e_1^i, \ \dot{\tilde e}_1^k &= -\Gamma^k_{ij} \tilde e_1^j \tilde e_1^i + \sum_{i=2}^d \tilde e_i^k \dot h^i, \ \dot{\tilde e}_l^k &= -\Gamma^k_{ij} \tilde e_l^j \tilde e_1^i – \tilde e_1^k \dot h^l, \quad 2\le l \le d. \end{aligned} \right. \end{equation} We have \begin{equation} \ddot{\tilde x}^i = \dot{\tilde e}_1^i = -\Gamma^i_{kj} \tilde e_1^j \tilde e_1^k + \sum_{k=2}^d \tilde e_k^i \dot h^k = -\Gamma^i_{kj} \dot{\tilde x}^j \dot{\tilde x}^k + \sum_{k=2}^d \tilde e_k^i \dot h^k, \end{equation} that is, \begin{equation}\tag{2*} \frac{\nabla \dot{\tilde x}^i}{dt} = \sum_{k=2}^d \tilde e_k^i \dot h^k. \end{equation} If (1*) and (2*) are the same ODE, then under the same initial condition $ x(0) = \tilde x(0) = x_0$ , we have $ x=\tilde x$ , which proves the claim. But I do not know how to compare (1*) and (2*).

Can anyone give some hints or reference? TIA…

PS: This is a crosspost from math.stackexchange.

Invariance under tame almost complex structure of the fibre tangent space of the symplectic normal bundle

I am trying to understand the construction of symplectic inflation and I am stuck in the following point.

Suppose we have a 4 dimensional symplectic manifold $ (M, \omega)$ . Also suppose that $ N \subset M$ is a 2 dimensional symplectic submanifold. Let $ J$ be a tame (with respect to $ \omega$ ) almost complex structure. Further we are given that $ N$ is $ J$ -holomorphic for the above $ J$ .

Let $ S(N)$ denote the symplectic normal bundle of $ N$ . Given a point $ p$ on the the intersection of a fibre $ F_p$ of $ S(N)$ and the zero section are the following statements true?

1)The tangent space $ T_p F_p$ is invariant under $ J$ .

2)Let $ w=(u,v) \in T_pF_p \oplus T_p N$ and $ w^\prime= (u^\prime,v^\prime)\in T_pF_p \oplus T_p N$ . Then $ \omega_p((u,v),(u^\prime,v^\prime)) = u^T J_p v + {u^\prime}^T J_p v^\prime$ .

I can see that these statements should be true when $ J$ is compatible with $ \omega$ , but I’m unable to show them for a tame almost complex structure.

A vector bundle associated to a codimension $1$ submanifold of a symplectic manifold

We consider the standard symplectic structure $ \omega=\sum dx_i\wedge dy_i$ on $ \mathbb{R}^{2n}$ . To every codimension $ 1$ submanifold $ M\subset \mathbb{R}^{2n}$ we associate a vector bundle $ E$ over $ M$ as follows:

$ $ E=\{(x,v)\in M\times \mathbb{R}^{2n}\mid \omega(v,N_x)=0,\;\;N_x\perp T_x M\}$ $

Does the structure of this bundle $ E$ , or at least its characteristic classes, depend on the way $ M$ is embedded in $ \mathbb{R}^{2n}$ ? What is the structure of this bundle for the standard odd dimensional spheres?Is it trivial?is it isomorphic to the tangent bundle of sphere?

One can extend this question by replacing $ \mathbb{R}^{2n}$ with an arbitrary symplectic manifold.

HKT manifolds with non trivial canonical bundle

A compact HKT manifold is a hyperhermitian manifold $ (M,I,J,K,g)$ such that either $ \partial (\omega_J+i\omega_K)=0$ (if endomorphisms act on the left on the tangent space and $ \partial$ is taken with respect to $ I$ ) or, equivalently, the three Bismut connections for hermitian structures $ (I,g)$ , $ (J,g)$ , $ (K,g)$ are all the same.

My question is related to the situation when the canonical bundle (with respect to $ I$ ) is non trivial$ ^1$ (holomorphically). Can it happen that $ \mathcal{K}_M$ has many section or it always has no sections at all, like for example for Hopf surfaces?

The second question is whether the answer is know for example for homogeneous examples of hypercomplex manifolds due to Joyce, $ SU(3)$ for instance?

$ 1.$ From the work of M. Verbitsky it is know that a compact HKT manifold has a trivial canonical bundle iff the holonomy of the Obata connection is reduced to $ Sl_n(\mathbb{H})$ .

Map to a given vector bundle from a split vector bundle

Let $ X$ be a connected scheme, smooth and proper over $ \mathbb{C}$ . Let $ F$ be a locally free $ \mathcal{O}_X$ -module of finite rank $ r>1$ . Suppose on a non-empty affine open $ U\subset X$ whose complement is irreducible we have an isomorphism of $ \mathcal{O}_X$ -modules $ f:\mathcal{O}^{\oplus r}_X|_{U}\rightarrow F|_{U}$ .

Does there necessarily exist a locally free $ \mathcal{O}_X$ -module $ L$ of rank $ 1$ with an identification $ \mathcal{O}_X|_U\rightarrow L$ such that there is a morphism of $ \mathcal{O}_X$ -modules $ g:L^{\oplus r}\rightarrow F$ with $ g|_{U}=f$ ?