mg.metric geometry – Geometric metric interpretation

For metric in $ mathbb R ^ 2 $ (polar coordinates) The elements of the Pythagorean triangle can be visualized in a differential form, as shown in the yellow triangle at the top. This happens for the metric:

$$ ds ^ 2 = r ^ 2 + (r d theta) ^ 2 $$
Applying the Euler-Lagrange variation calculation equation, we have DE and a particular integral curve (initiated on $ theta) $

$$ r ^ 2 + 2r {2} -r r ^ {& # 39; & # 39;} = 0, quad r = sec , theta $$
We have two sides for $ r d theta, dr $ and a hypotenuse, these are well known.

Consider the metric

$$ ds ^ 2 = r ^ 2 – (r d theta) ^ 2 $$

The DE and its integral curve solution turn out to be

$$ r ^ 2 – 2r {2} + r r ^ {& # 39; & # 39;} = 0, quad r = sech , theta $$

and they are shown in the little yellow triangle on the left.

Interpretation of metric surveyors

However, the third side is not seen shorter $ r = sech , theta $ curve but still Remains like the longer hypotenuse as for the metric case of Pythagoras.

How can we geometrically interpret the sides / hypotenuse directly reflecting the metric?

Geometric topology – Does each homogeneous metric space connected locally compact admit a green-transitive "grid"?

The answer is no. A quick response can be made as follows:

(1) Pansu proved (1989) that two Carnot Lie groups are quasi-isometric if and only if they are isomorphous.

(2) There is a continuum of many non-isomorphic 7-dimensional Carnot Lie groups.

Yes $ Y $ is appropriate, uniformly discrete and isometric-transitive, then he is IQ at his isometric group $ G $which is locally compact. Assuming that QI to a nilpotent Lie group further implies (by the results of Gromov / Losert / Trofimov) that $ G $, modulo a compact normal subgroup, is discrete and practically nilpotent. There are only many classes of quality assurance. Therefore, at least one of the examples in (2) gives (for each choice of left invariant Riemannian metric) a negative answer to your question.

There is an alternative using hyperbolicity rather than polynomial growth, consisting of a continuum family of 3-fold homogeneous curved negatively. That they are not IQ is also the result of Pansu. The fact that a totally local compact IQ group must be compact by discrete is an immediate consequence of Hilbert-Smith's two-dimensional conjecture (which is an ancient theorem).

Real Analysis – Not quite a series: disrupting a geometric sequence

Let $ alpha in mathbb {R}: 0 < alpha <1 $. Then the sequence
$$
sum_ {k = 0} ^ n binom {n} {k} (- alpha) ^ k
$$

converges to zero (applies the binomial theorem). What about the following? Say, $ (x_n) $ is a bounded sequence of complex numbers, and we take
$$
a_n: = sum_ {k = 0} ^ n binom {n} {k} (- alpha) ^ k x_k.
$$

Under what circumstances, if any, can we conclude that
$$
| a_n | leq | Omega_1 (1- alpha) ^ n + Omega_2 |
$$

for some people $ Omega_1, Omega_2 $, for all but a lot $ n $? Under what circumstances can we expect convergence? Thoughts? Thank you.

python – Creates a geometric automaton

I'm trying to draw 2D patterns on grids of different geometric units. I have a small example for a half-moon model on the grid below. In the future, I will add different types of movement (for example, a complete or circular motion). Finally, there will also be different basic geometric units.

I wonder if there is a way to clean the direction code because it is big enough and splashed randomly. I also wonder if there is a better way to write the initialization of the grid itself (which is currently a loop for j i).

All other tips would also be appreciated.

import tkinter as tk
import time
import numpy

# set hex colors
cream = '#fafaeb'
umber = '#21201f'

root = tk.Tk()
canvas = tk.Canvas(root, width=650, height=450, bg=umber)
canvas.pack(fill="both", expand=True)

# TODO: freezes when trying to click again sometimes

direction='DOWN'
# note that tkinter canvas starts with an origin in the corner, moving down therefore is *adding* y value,
# so some of the math may look inverted

motion='HALFMOON'

# change direction or motion
def apply_directives(event):
    global direction
    global motion
    if event.char == 'u':
        direction = 'UP'
    if event.char == 'd':
        direction = 'DOWN'
    if event.char == 'r':
        direction = 'RIGHT'
    if event.char == 'l':
        direction = 'LEFT'

    if event.char == 'c': # stands for 'click'
        motion = 'CLICK'

root.bind("", apply_directives)



WIDTH = 6



# apply bold to line
def bold(event):
    # initial boldness
    # find arc user meant
    id = event.widget.find_closest(event.x,event.y)(0)
    # retrieve arc tag
    tag = canvas.gettags(id)(1)
    # bold arc
    canvas.itemconfigure(id,width=WIDTH)
    # redraw canvas
    canvas.update()
    # give time to make each drawing piecemeal
    time.sleep(.5)

    if motion == 'HALFMOON':

        # find within the next enclosed box in the right, the arc with a tag that fits the motion type so long as
        # there are no more arcs to the right
        set_a = 0
        set_b = 0

        # sets inverse depending on vertical or horizontal direction
        if direction == 'RIGHT' or direction == 'LEFT':
            set_a = ('1','2')   # tags are in type string, so we match the type
            set_b = ('3','4')
        if direction == 'UP' or direction == 'DOWN':
            set_a = ('1','4')   # tags are in type string, so we match the type
            set_b = ('2','3')

        current_set = ()

        # check to see what kind of curve this is
        if tag in set_a:
            current_set = set_a
        else:
            current_set = set_b

        # TODO: Sometimes takes an arc that shouldn't be within the bounding box, but can't consistently
        #  replicate this. find out way and fix
        # possibly when you double click?

        # direction logic
        directional_additive = 0
        if direction == 'RIGHT':
            directional_additive = numpy.array((arc_width,0))
        if direction == 'LEFT':
            directional_additive = numpy.array((-arc_width,0))
        if direction == 'UP':
            directional_additive = numpy.array((0,-arc_width))
        if direction == 'DOWN':
            directional_additive = numpy.array((0,arc_width))
        prev_id = 0
        # TODO: buggy while. figure out previous id while loop?
        # when there are no more arcs to the desired direction
        while (id != prev_id):
            print('in while')
            print('x is ' + str(event.x))
            print('x bound is ' + str(canvas.winfo_width() - arc_width))
            print('y is ' + str(event.y))
            print('y bound is ' + str(canvas.winfo_height() - arc_width))
            # set up variables to find next coordinates
            current_box_coords = numpy.array(canvas.coords(id))
            # box is too big, we just want the arc box
            normalizer = 0
            if tag == '1': # take upper right of box
                normalizer = numpy.array((arc_width,0,0,-arc_width))
            if tag == '2': # take upper left of box
                normalizer = numpy.array((0,0,-arc_width,-arc_width))
            if tag == '3': # take lower left of box
                normalizer = numpy.array((0,arc_width,-arc_width,0))
            if tag == '4': # take lower right of box
                normalizer = numpy.array((arc_width,arc_width,0,0))
            current_arc_coords = current_box_coords + normalizer
            next_coords_additive = 0
            # directional logic
            if direction == 'RIGHT':
                next_coords_additive = numpy.array((arc_width,0,arc_width,0))
            if direction == 'LEFT':
                next_coords_additive = numpy.array((-arc_width,0,-arc_width,0))
            if direction == 'UP':
                next_coords_additive = numpy.array((0,-arc_width,0,-arc_width))
            if direction == 'DOWN':
                next_coords_additive = numpy.array((0,arc_width,0,arc_width))
            # tkinter's find_enclosed method will exclude any objects it finds right at the perimeter, so make the perimeter slightly larger
            boundaries_additive = numpy.array((-1,-1,1,1))
            # obtain the next coordinates
            next_coords = current_arc_coords + next_coords_additive + boundaries_additive

            # obtain list of the next IDs
            next_ids = event.widget.find_enclosed(*next_coords)

            # obtain list of the next tags
            next_tags = (canvas.gettags(next_id)(1) for next_id in next_ids)

            last_event_x = event.x
            prev_id = id

            for next_id,next_tag in zip(next_ids,next_tags):
                if ((id != next_id) & (next_tag in current_set)):
                    # move cursor to the desired direction
                    if direction == 'RIGHT':
                        event.x += arc_width
                    if direction == 'LEFT':
                        event.x -= arc_width
                    if direction == 'UP':
                        event.y -= arc_width
                    if direction == 'DOWN':
                        event.y += arc_width

                    # bold the new arc
                    canvas.itemconfigure(next_id, width=WIDTH)
                    canvas.update()
                    time.sleep(.5)
                    # update current arc
                    id = event.widget.find_closest(event.x, event.y)(0)
                    # update current tag
                    tag = canvas.gettags(id)(1)

# each bounding box is 100 x 100
class Box():
    def __init__(self, coords):
        # give the class a tag for tkinter to find later
        self.tag = 'box{}'.format(id(self))

        # make each arc
        self.arcs = (
            # arc 1
            canvas.create_arc(coords, start=0, extent=90, outline=cream, style="arc", tag=(self.tag, 1)),
            # arc 2
            canvas.create_arc(coords, start=90, extent=90, outline=cream, style="arc", tag=(self.tag, 2)),
            # arc 3
            canvas.create_arc(coords, start=180, extent=90, outline=cream, style="arc", tag=(self.tag, 3)),
            # arc 4
            canvas.create_arc(coords, start=270, extent=90, outline=cream, style="arc", tag=(self.tag, 4))
        )

        # allow each arc to be bolded
        self.bind()

    def bind(self):
        # apply binding to every arc in box
        for arc in self.arcs:
            canvas.tag_bind(arc, "", bold)

# coordinates are (x,y) of upper left corner, and then (x,y) of lower left corner
# use numpy array for vector addition
coords = numpy.array((0, 0, 100, 100))


# use box width to calculate grid indice
box_width = coords(2) - coords(0)
# grid indice to move around
grid_indice = box_width/2

# use arc width for width of 1 component
# 4 components in 1 box
arc_width = box_width/2

# make desired size of grid (width, height)
size=(6*2,4*2)

for i in range(size(1)):
    for j in range(size(0)):
        # keep adding 1 grid indice to the x as you move to the right
        box_coords = coords + numpy.array((0 + grid_indice*j, 0, 0 + grid_indice*j, 0))

        # create variables to check parity
        even_row = i%2 == 0
        odd_row = not even_row
        even_column = j%2 == 0
        odd_column = not even_column

        # only draw a box on the same parity of i and j
        # that is: on an odd row (i), only draw on odd column (j) values
        if even_row & even_column:
            Box(tuple(box_coords))
        elif odd_row & odd_column:
            Box(tuple(box_coords))

    # keep adding 1 grid indice to the y as you move down
    coords = coords + numpy.array((0, 0 + grid_indice, 0, 0 + grid_indice))

root.mainloop()

```

Differential geometry – Geometric interpretation for a uniformly elliptic pde of second order

Let $ Omega subset mathbb {R} ^ {2} $ a domain, let $ u in C ^ {2} ( Omega) $, the operator

$ Lu = tr (A.D {{2} u) + + cu $

or $ A $ is a symmetric matrix, $ b $ is a continuous vector field on $ mathbb {R} ^ {n} $ and $ c in C ( Omega) $ is uniformly elliptical if there are constants $ lambda, Lambda $, such as $ lambda leq Lambda $ where, for all $ xi in mathbb {R} ^ {n} $

$ lambda | xi | ^ {2} leq xi ^ {t} A xi leq Lambda | xi | ^ {2} $ .

What is the relationship between the control of the underlying values ​​and the equation? What is the geometric interpretation of the uniformly elliptical equation? There is a relationship with curvature?

For me, being uniformly elliptical means having an ellipse trapped between two paraboloids, one with a smaller aperture and the other with a larger aperture.

reference request – geometric morphism whose country is epic

Is there a name for the class of geometric morphisms whose advice is epic (equivalent, whose direct images are faithful)?

This includes the class of inclusions / incorporations of topos. For example A4.6.2 in Johnstone & # 39; s Sketches of an Elephant: Compendium of the theory of toposAll of these morphisms are local, but since Johnstone does not include a name (and its cross-references are usually very deep), I wait for them not to appear in the text.

If no one can dig up a reference, I would like to highlight injection as a name for this property, not only because fidelity gives injectivity on hom-sets, but also because it is the conceptual dual of a geometric surjection, which has a monical unity. faithful inverse image.

Geometric interpretation of (anti) derivatives

Integration is anti-derivative and conversely, analytically. But geometrically, the integration corresponds to the area under the curve, while the derivative is the slope of the curve at a given point. How to understand the difference? Geometrically, how are they similar?
0
P. I come from physics.

Geometric Glovius Pro 5.1.0.428 – NulledTeam UnderGround

Size of the file: 191 MB

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View the product structure, hide / show, move, and search for components.
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View and search attributes, export of the BOM table.
Take accurate linear, angular, and radial measurements.
Take sections with several cutting planes, comb and describe the support.
Compare any two models. Visualize the differences in 3D.
Export to PDF 3D, 3DS, OBJ, STL and Glovius Mobile.

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dg.differential geometry – Elementary geometric proof of Hirzebruch's signature theorem for Riemannian varieties of 4?

Hirzebruch's signature theorem tells us that for a compact smooth-oriented variety, the signature $ sigma (M) $ is proportional to the first number of Pontryagin's $ M $:
$$
3 sigma (M) = p_1 (M) = k int_M mathrm {tr} (R ^ 2)
$$

for an appropriate constant k $ (For the purpose of this question, consider the integral as the definition of $ p_1 $, although this correctly uses the Chern – Weil theory to prove it). The usual proof of this fact is a general theorem involving the $ L $-kind. The question has recently been raised in conversations about whether this result had a more basic geometric proof, for example at the intrinsic proof level of the Gauss-Bonnet theorem, with additional specializations specific to dimension 4.

Some simple observations:

  1. It can (apparently) simplify the integral to one of the forms $ int_M | W ^ + | ^ 2 – | W ^ – | ^ 2 $ (up to a proportionality constant), using the decomposition of the curvature tensor $ R $ via the Hodge theory and the auto and anti-auto-double parts of the Weyl curvature, $ W ^ pm $.

  2. Similarly, we can write the signature as difference $ b_ + – b _- $ dimensions of a decomposition of $ H ^ 2 $.

So, we might be able to prove $ 3 (b_ + – b_-) = frac {1} {4 pi ^ 2} int_M | W ^ + | ^ 2 – | W ^ – | ^ 2 $ more directly using geometric methods, but I am absolutely unfamiliar with this field and existing research. We can clearly prove the result by using the same pigeon-fancier methods, but that's not what I'm looking for.

A geometric property of some Lie groups

What is an example of a Lie group $ G $ not ismorphe to the upper half of Poincaré $ H ^ n $ but satisfy the following:

For each left invariant metric $ g $ we draw all the geodesics. Then we can resize $ g $ conform to a metric $ h $ and we observe that at every point $ x in G $
all walnut half-radius $ g $– initiation of geodesics $ x $ to have finished $ h $-length except a geodesic tangent to a single direction