Do you die on another plane if you reached that plane via Astral Projection?

Inspired by this question.

I’m not sure that I understand how the Astral Projection spell works. The description states the following:

You and up to eight willing creatures within range project your astral bodies into the Astral Plane (the spell fails and the casting is wasted if you are already on that plane). The material body you leave behind is unconscious and in a state of suspended animation; it doesn’t need food or air and doesn’t age.

Your astral body resembles your mortal form in almost every way, replicating your game statistics and possessions. The principal difference is the addition of a silvery cord that extends from between your shoulder blades and trails behind you, fading to invisibility after 1 foot. This cord is your tether to your material body. As long as the tether remains intact, you can find your way home. If the cord is cut—something that can happen only when an effect specifically states that it does—your soul and body are separated, killing you instantly.

Your astral form can freely travel through the Astral Plane and can pass through portals there leading to any other plane. If you enter a new plane or return to the plane you were on when casting this spell, your body and possessions are transported along the silver cord, allowing you to re-enter your body as you enter the new plane. Your astral form is a separate incarnation. Any damage or other effects that apply to it have no effect on your physical body, nor do they persist when you return to it.

The spell ends for you and your companions when you use your action to dismiss it. When the spell ends, the affected creature returns to its physical body, and it awakens. The spell might also end early for you or one of your companions. A successful dispel magic spell used against an astral or physical body ends the spell for that creature. If a creature’s original body or its astral form drops to 0 hit points, the spell ends for that creature. If the spell ends and the silver cord is intact, the cord pulls the creature’s astral form back to its body, ending its state of suspended animation.

If you are returned to your body prematurely, your companions remain in their astral forms and must find their own way back to their bodies, usually by dropping to 0 hit points.

My specific point of confusion, I think, centers around the difference between the terms ‘astral form’, ‘original body’, and ‘physical body’. Their intent is not clear to me as the spell is written.

My interpretation of the spell is that it permits you to travel the astral plane to locate other planes of existence in an astral form, which essentially your soul and mind. You leave behind your physical body with a silver cord tethering it to your astral form.

If you drop to 0 hit points during astral travel, you are safe because your astral form can follow the silver cord to rejoin with your physical body and thus not suffer significant consequences from failed exploration. Upon finding a portal to another plane, you can enter it and your physical body is pulled along the silver cord to rejoin with your astral form (i.e. mind) and thereby allowing you to explore that plane.

Is this combined form your physical body, astral form, or something else?

Is this interpretation correct? Other answers suggest that you can travel to other planes in a manner similar to summoned creatures and thus avoid some of the dangers of death by virtue of having an automatic retreat to your physical body, however, this doesn’t make sense if your physical body is pulled along the cord to rejoin with your astral form. The spell states that the cord is between your physical body and astral form, I would think that once your body was pulled along the cord, the cord would effectively not exist.

I think my interpretation may not be as generous as others especially given that this is a 9th level spell, but I had based it upon the notion that the Astral Plane was a different and more unusual plane than others, which required special consideration to navigate in a ‘safe’ manner.

Constraint propagation using Projection rule

I’ve found this example of constraint propagation using projection rule enter image description here

We have

C = { x1 ≠ x2, x1 ≥ x2 }  < C; x1 ∈ {1,2,3}, x2 ∈ {1,2,3} > 

They say that applying propagation rule, does not give any simplification.

I’m not sure why this is the case. Shouldn’t we get?

< C; x1 ∈ {2,3}, x2 ∈ {1,2} > 

Other steps in the example, make sense that to me, e.g.

< C; x1 ∈ {2}, x2 ∈ {1,2,3} > 


< C; x1 ∈ {2}, x2 ∈ {1} > 

How to use known Projection Matrices of cameras to generate new fundamental matrix located correctly in 3D space?

I have a few cameras which have been calibrated (using a checkerboard) so I know the fundamental matrix between an origin camera and each remaining camera

I wish to take pairs of camera with no fundamental matrix – but each has a projection matrix – and calculate the fundamental matrix with the view to reconstruct the stereo pair in 3D. To clarify – as all cameras have a projection matrix – I should be able to take any combination and generate a Fundamental matrix – so I can reconstruct and combine stereo pairs in 3D using as many pairs as possible

currently I use this formula: enter image description here

(decomposing Projection matrices from camera to get C)

but that didn’t quite work with stereo pairs that don’t include the origin. I noticed that to calculate the fundamental matrix that rotation was relative so I added the relative rotation formula when recalculating P from K/R/T (intrinsic camera params, Rotation matrix, Translation matrix)

enter image description here

but now each reconstruction appears on the camera position (I think) instead of all stereo pairs being mapped into same space to give a dense reconstruction

Does anyone know what I am doing wrong? thanks for any help

Graph of function, continuous projection

$ X$ and $ Y$ are topological spaces. $ f:X\rightarrow Y$ a map (we don’t suppose that $ f$ is continuous). Consider $ A=\{(x,f(x))\in X\times Y| x\in X\}$ . is $ \pi: A\rightarrow X$ , $ $ (x,f(x))\mapsto x$ $ a homeomorphism ? If not, is it enough to assume that $ A$ is closed subspace in $ X\times Y$ ? If $ X$ and $ Y$ are metrizable spaces, how to prove that $ \pi$ is a homemorphism using sequential continuity ? Suppose that by miracle $ \pi$ is homeomorphism but $ f$ is not continuous, is it possible such phenomenon ?

Projection of a polytope along 4 orthogonal axes

Consider the following problem:

Given an $ \mathcal{H}$ -polytope $ P$ in $ \mathbb{R}^d$ and $ 4$ orthogonal vectors $ v_1, …, v_4 \in \mathbb{R}^d$ , compute the projection of $ P$ to the subspace generated by $ v_1, …, v_4$ (and ouput it as an $ \mathcal{H}$ -polytope).

I know that the problem of computing projections along $ k$ orthogonal vectors in NP-hard (if $ k$ and $ d$ are part of the input), as shown in this paper. But does it help if $ k$ is a constant? Specifically, does it help if $ k \leq 4$ ? Do we have a polynomial algorithm in this case?

Estimating quality of projection

I asked this question on, but didn’t receive an answer

Suppose we are given a vector $ v$ and vectors $ \mu_i$ :

$ v = \mu_1+\mu_2+…+\mu_m$ , where $ \mu_i \in R^n$ , all $ \mu_i$ are of unit length.

Oracle will give me $ k$ vectors $ \mu_{j_1}, \mu_{j_2},…\mu_{j_k}$ from the original set such that when I project $ v$ onto subspace spanned by these vectors the length of the projection is highest possible. In other words, from the set of all combinations of $ k$ vectors from $ [\mu_1,…\mu_n]$ the $ [\mu_{j_1}, \mu_{j_2},…\mu_{j_k}]$ give highest length of projection. Lets denote by $ v_{\text{proj}}$ projection of $ v$ onto $ [\mu_{j_1}, \mu_{j_2},…\mu_{j_k}]$

I want to estimate quality of projection before oracle gives me this $ k$ vectors. I want to give upper bound on $ ||v – v_{\text{proj}}|| $

As far as I understood it is very difficult to obtain these $ k$ vectors by myself. However, I know that for any two vectors $ \mu_i, \mu_j$ , $ ||\mu_i-\mu_j|| \leq \alpha$ , where $ \alpha$ is a given positive number.

Small values of $ \alpha$ will tell me that all $ \mu_i$ are close to each other and heading towards same direction. I would suspect then that projection will be good, and its length will be close to the length of original vector. How can I use this to give an upper bound $ ||v – v_{\text{proj}}|| $ ?

My attempts:

Without loss of generality lets assume that $ k$ optimal vectors are first $ k$ vectors in the list, i.e $ \mu_1,\mu_2,…\mu_k$ . Lets denote by $ P$ projection operator on the space spanned by $ \mu_1,\mu_2,…\mu_k$ .

$ \|v – v_{\text{proj}}\| = \|v – P(v)\| = \|v – P(\mu_1+\mu_2+…+\mu_m)\| = $

$ \|v – P(\mu_1) – P(\mu_2) – … – P(\mu_m)\| = $

$ \| v – \mu_1 – \mu_2 – … – \mu_k – P(\mu_{k+1}) – P(\mu_{k+2}) – … – P(\mu_m)\| = $

$ \|\mu_{k+1} – P(\mu_{k+1}) + \mu_{k+2} – P(\mu_{k+2}) + … + \mu_{m} – P(\mu_{m})\|$

$ \|v – v_{\text{proj}}\| \leq \|\mu_{k+1} – P(\mu_{k+1})\| + \|\mu_{k+2} – P(\mu_{k+2}) + … + \|\mu_{m} – P(\mu_{m})\|$

$ \|v – v_{\text{proj}}\| \leq (m-k)\alpha$

So in order to make $ \|v – v_{\text{proj}}\| \leq \epsilon$ , we need $ k \geq \frac{m\alpha – \epsilon}{\alpha}$

I am not satisfied with this result because $ k$ grows linearly with $ m$ . I want it to grow much slower, something like $ \log(m)$ . My goal is to show that under some constraints on $ \mu_i$ , we need only approximately $ \log(m)$ vectors to approximate $ v$ .

I think the bound can be improved substantially. First Cauchy inequality isn’t very tight and second, I used $ |\mu_{k+1} – P(\mu_{k+1})\| \leq \alpha$ which is also very loose.

I am open for additional constraints on $ \mu_1,…\mu_m$ to achieve logarithmic growth

Alex Ravsky on has noted, that we also need a constraint on $ \alpha$ in order to achieve logarithmic growth. Assume that $ k$ $ \leq n$ , $ \mu_i$ is th $ i$ -th standard ort of the space $ \mathbb{R}^n$ , and $ \alpha = \sqrt{2}$ . Then $ \|v – v_{\text{proj}}\| = \sqrt{m-k}$

Projection of an invariant almost complex structure to a non integrable one

My apology in advance if my question is obvious or elementary

We identify elements of $ S^3$ with their quaternion representation $ x_1+x_2 i +x_3 j +x_4 k$ . We consider two independent vector fields $ S_1(a)=ja$ and $ S_2(a)=ka$ on $ S^3$ . On the other hand $ P: S^3\to S^2$ is a $ S^1$ -principal bundle with the obvious action of $ S^1$ on $ S^3$ . Then the span of $ S_1, S_2$ is the standard horizontal space associated to the standard connection of the principal bundle $ S^3 \to S^2$ . Then each horizontal space has an almost complex structure $ J$ . This is the standard structure associated to $ S_1, S_2$ coordinate.

Is this structure invariant under the action of $ S^1$ ? If yes, we can define a unique almost complex structure on $ S^2$ which is $ P$ related to the structure on total space. Now is this structure on $ S^2$ integrable?

As a similar question, is there an example of a principal bundle $ P\to X,$ such that $ P$ is a real manifold and $ X$ is a complex manifold and a connection admit an invariant almost complex structure which project to a non integrable structure?

Uniqueness of projection under spectral norm

I am considering $ $ \min_{M\in \mathcal{M}} \|X – M\|:=x \neq 0, $ $ where $ X$ , $ M$ are $ m\times n$ matrices, $ \|\cdot\|$ is spectral norm and $ \mathcal{M}$ is a matrix subspace. I wonder to what extent the solutions are unique, i.e. what can we say about the set $ $ \mathcal{M}^* = \{M|\|X – M\|=x\}. $ $ Currently, I only know it is a convex set. For $ M_1, M_2 \in \mathcal{M}^*$ , let $ M = p M_1 + (1-p) M_2$ , $ p \in (0,1)$ . For any vector $ v$ , using Cauchy’s inequality, we have $ $ v^T(X – M)^T(X – M)v \leq x^2, $ $ thus $ M \in \mathcal{M}^*$ . Moreover, the equality holds only when $ $ (X – M_1)v = (X – M_2)v = u,~~~~s.t.~u^T u = x^2. $ $ I wonder whether there are more properties I could say about this set, e.g. its dimension/the properties of the common eigenvectors/the condition under which the solution is unique.

Projection on space of permutations of lower diagonal matrices

Let $ G$ be the group of all $ n\times n$ permutation matrices and let $ V$ be the vector space of all $ n$ -dimensional lower diagonal matrices. Then I define the set $ $ X = \{P\cdot L\cdot P^T \mid \forall P\in G, L\in V\}, $ $ where “$ \cdot$ ” is matrix multiplication. That is, $ X$ is the set of all lower diagonal matrices with possible rows and colums that are permuted at the same time.

I am interested in the orthogonal projection of a general matrix $ M\in\mathbb{R}^{n\times n}$ on the set $ X\subset \mathbb{R}^n$ . Does anyone know if there is a more efficient method to do it than projecting on each space $ $ V_P = \{P\cdot L\cdot P^T \mid \forall L\in V\} $ $ for every individual $ P\in G$ and then taking the shortest one?

Thank you in advance!