I am trying to understand how the orientation is induced on the boundary of a differentiable manifold with boundary. Here is what I have worked out so far:  Let $M$ be a differentiable manifold and let $\partial M$ denote its boundary. Let $U$ be an open set in $M$ with $U \cap \partial M \neq 0$. Similarly let $V$ be an open set in $M$ with $V \cap \partial M \neq 0$. Let $x_i$ and $y_i$ be local coordinates in $U$ and $V$ respectively. On the boundary $x_1 = 0 = y_1$. Furthermore we know that the function $y_i = y_i (x_1,\dots, x_m)$ is differentiable. Then the orientation on $\partial M$ will be the orientation given by $x_i$ and $y_i$. It remains to be shown that this is well-defined that is the same on $\partial M \cap U \cap V$. Assuming what I wrote so far is correct my question would be how to do this final step?
Given a $k$-manifold $M$, such that $\partial M$ is a $(k-1)$-manifold, there is a standard way in which $\partial M$ inherits the orientation of $M$. So if $M$ is oriented by the form field $\omega$, then $\omega_\partial=\omega(v_{\text{out}},v_1,\dots,v_{k-1})$ orients $\partial M$, where the $v_1,\dots,v_{k-1}$ are basis vectors for a tangent space on a point of the boundary, and $v_\text{out}$ is an outward pointing vector (see here: , which follows the exposition of Hubbard &amp; Hubbard).  In practice I find this very difficult to interpret. How can one find this $v_\text{out}$ in practice?   For instance, take $M$ to be defined by $z=x^2+y^2$ for $x\ge0$ (technically a manifold with boundary, in fact a half-paraboloid). It's a fact that $\partial M$ is defined by the parabola $z=y^2$. If $M$ is oriented by $dx\wedge dy$, how does this induce an orientation on the boundary (in fact we should obtain a 1-form that induces an orientation on the boundary)?   Note: I find the whole discussion of boundary orientation to be rather clumsy in its exposition. In the case of surfaces inducing orientations on boundary curves, I was originally taught to imagine that the normal vector that induces the surface orientation is me walking on the surface. This normal should walk along the boundary such that the surface lies to its left. OK - but how does this translate to the $v_{\text{out}}$ interpretation, so that we can generalize to other dimensions easily?)  Edit: An ideal answer would merge the intuitive approach with the rigorous approach, if possible. I would grateful to anybody who could help order my (at present confused) understanding!
Let $\Gamma$ be a circle that passes through the origin. Show that we can find real numbers $s$ and $t$ such that $\Gamma$ is the graph of $r = 2s \cos (\theta + t).$  I know this has to be converted to a cartesian equation, but how do I do this, and what do I do after?  Thanks Why doesn't C3PO recognize Leia in R2's recorded message? sudo find / -name "*" | xargs grep -sn --color=auto "-j"   Above command returns below:  grep: invalid option -- 'j' Usage: grep [OPTION]... PATTERN [FILE]... Try `grep --help' for more information. ...   How do I search for string -j? I am currently studying  answer on the Moore-Penrose pseudoinverse and the Euclidean norm by the user &quot;Etienne dM&quot;. The first point of their answer proceeds as follows:  Let $x$ be $A^+y$.  Let me begin by the second point. For all $z$, we have: \begin{align} \lVert Az-b \rVert_2^2 &amp;= \lVert Ax-b \rVert_2^2 + \lVert A(z-x) \rVert_2^2 + 2 (z-x)^TA^T(Ax-b)\\ &amp; \geq  \lVert Ax-b \rVert_2^2 + 2 (z-x)^TA^T(Ax-b) \end{align} Moreover, because $(AA^+)^T = AA^+$, $$ A^T(Ax-b) = ((AA^+)A)^Tb - A^Tb = 0$$ Thus, we prove that for all $z$, $\rVert Az-b \lVert_2^2 \geq\rVert Ax-b \lVert_2^2$, that is to say $A^+b$ is as close as possible to $y$ in term of the Euclidian norm $\lVert Ax-b\rVert_2$.   I realise that $x = A^+ y$, but I don't understand how any of this implies that &quot;$A^+ b$ is as close as possible to $y$ in terms of the Euclidean norm $\lVert Ax-b\rVert_2$&quot;. I would greatly appreciate it if people would please take the time to explain this to me. What, exactly, is transferred along supply lines? After learning about the  datasets I've working with the following layers, from the USA data   Country boundaries State boundaries County boundaries City boundaries ZipCode/Postcode boundaries   What I am trying to find is the equivalent (where appropriate) of the same data for various countries.  Is there any free administrative boundaries available as shapefiles for other countries? I dragged a file to the recycle bin a few days ago. I have Windows 7 premium on a 245 GB partition, a 240 GB SSD, and a 250 SSD. And somewhere in the past few days I incidentally installed Delta Search. I know I should remove it right away, but I just switched to Windows on the partition of a 3.5 1 TB drive and would like to restore that file and deal with the malware after Christmas is over.  I'm pretty sure I know what drive that deleted files should be on, but what is the real name of the recycle directory and exactly where is it located? 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IE the vertical component of moving 400lbs at a 45 degree angle. Vote too old to be changed, unless post is edited Anyone seen this one yet? Vote too old to be changed, unless post is edited My vote was only a few hours old. Then I realized the answer was wrong. Updated my question and went to remove the vote. Even after updating question I still have the issue. Small issue, but frustrating.  See also: I need to build a probabilistic cellular automaton for simulating land-use changes and am wondering how to implement this in a GIS. I understand that Markov chains are a way of doing this, and IDRISI has some sort of module for Markov chains, but I have no access to IDRISI.  that, with raster data, this can, in theory, be done with any GIS that can do raster algebra. Presumably this means that it can even be done in ArcGIS, and I think I've figured out how to build cellular automata in ModelBuilder (once I know with certainty how to make iteration work there).  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