User:KSmrq/Sandbox

Image tests


(See Help:Contents/Images and media.

From Picture tutorial:


 * But 'thumb' also automatically resizes a large picture into a smaller display size, with an option for the user to click on the image and see the original large version. Because different people work to different screen resolutions, your preferred size of thumbnails can be set in special:preferences under "files".  The default, which is also used for logged-out users is 180 pixels (px), but you can choose between 120px, 150px, 180px, 200px, 250px, and 300px sizes.  If an image is smaller than the thumbnail size you specified then it is displayed at 100% resolution, i.e. its natural width.  Generally speaking, thumbnails are the best way to display images.

So it looks like 180px is the default thumbnail width, with aspect ratio preservation giving height.



Spin and Pin
A subtle technical problem afflicts some uses of orthogonal matrices. Not only are the group components with determinant +1 and −1 not connected to each other, even the +1 component, SO(n), is not simply connected (except for SO(1), which is trivial). Thus it is sometimes advantageous, or even necessary, to work with a covering group of SO(n), the  spin group, Spin(n). Likewise, O(n) has covering groups, the Pin groups, Pin(n). For n > 2, Spin(n) is simply connected, and thus the universal covering group for SO(n). By far the most famous example of a spin group is Spin(3), often seen in the form of unit quaternions or Pauli spin matrices.

In peculiarly Ouroboros fashion, the Pin and Spin groups are found within Clifford algebras, which themselves can be built from orthogonal matrices. The matrices used must be large enough to accomodate the entire algebra, which has dimension 2n. Any signature is possible; for example, in Cℓ0,1 the sole basis vector e1 squares to −1, while in Cℓ3,0 the three basis vectors e1, e2, e3 all square to +1. In all cases 1 takes the form of an identity matrix. To illustrate the first case, use 2×2 matrices and let the vector basis be


 * $$J = \begin{bmatrix}0&-1\\1&0\end{bmatrix} .$$

This trivial case is nevertheless interesting, because Cℓ0,1 ≅ ℂ, with J acting like the imaginary unit, i: J2 = −I. The Cℓ3,0 case can immediately put this to work. Use 4×4 matrices partitioned into 2×2 blocks, where I is the 2×2 identity and J is — as above — the 2×2 equivalent of i. Take as vector basis the following.

Per signature, eα2 = 1 (the identity) for all α; and basis vector products anti-commute, eαeβ = −eβeα for all α ≠ β, as required for a Clifford algebra. Writing eαβ = eαeβ, use these anti-commuting products as a basis for bivectors.

The square of each of these matrices is −1, as is the product of all three, e32e13e21. These are the famous Brougham Bridge relations for quaternions, i'2 = j2 = k2 = ijk = −1, thus this bivector basis can be identified with the quaternion units: e32 ≅ i, e13 ≅ j, e21 ≅ k. Continuing the correspondence, each of these matrices is skew-symmetric while the identity is symmetric; thus transposition acts like quaternion conjugation.

Top the algebra with the triple product e123 = e1e2e3.




 * $$\mathbf{e}_{123} $$ || $$ = \begin{bmatrix}J&0\\0&J\end{bmatrix} $$
 * || $$ = \begin{bmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0\end{bmatrix} $$
 * }
 * || $$ = \begin{bmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0\end{bmatrix} $$
 * }

Including the 4×4 identity as 1, this completes a list of eight linearly independent matrices, and a basis for Cℓ3,0. (The particular choice of basis matrices used in this example is not inevitable; other variations exist obeying the same essential algebraic relations.) Scalars are real multiples of 1; vectors are real linear combinations of the eα; bivectors, of eαβ; pseudoscalars are real multiples of e321; and a general element is a linear combination of all. Inspection confirms that Cℓ3,0 ≅ ℂ2×2, the 2×2 complex matrices. Also, compare the bivector basis to the Pauli matrices.

Now consider the product eβeα(−eβ) for different values of α and β. When α = β the result is −eα; otherwise, eα. Replace eα by an arbitrary vector v, and extend from the basis by linearity; v is reflected, by negating its eβ component. More generally still, replace eβ by any unit vector u; then uv(−u) is a reflection of v, negating its u component.

In Euclidean geometry the composition of two reflections, unit normals u1 and u2, is a rotation of the subspace the normals span, turning by twice the angle between them. (Any vector component orthogonal to both normals is unmoved.) Thus consider u1 = e2 followed by u2 = e3:


 * $$\mathbf{e}_3 \big(\mathbf{e}_2 \mathbf{v} (-\mathbf{e}_2)\big) (-\mathbf{e}_3) = (\mathbf{e}_3 \mathbf{e}_2) \mathbf{v} (\mathbf{e}_2 \mathbf{e}_3) = \mathbf{e}_{32} \mathbf{v} (-\mathbf{e}_{32}) $$

Since e2 and e3 are orthogonal, e32 produces a 180° rotation preserving their mutual perpendicular, the e1 axis. Another indication that this rotation is a half turn is that it is its own inverse, a consequence of the bivector on both sides of v squaring to −1. The same is true for any unit magnitude bivector; but if u1 and u2 are not perpendicular, then their product includes a scalar as well as a bivector component. For example, (1+e32)/√2 produces a 90° rotation preserving the e1 axis. Inclusion of scalars completes the construction of Spin(3). Negating a rotator has no effect on the rotation produced, so Spin(3) is a double cover of SO(3).

Inclusion of scalars also completes the correspondence with quaternion algebra; thus Spin(3) is isomorphic to the group of unit quaternions, or S3. Furthermore, inspection of the bivector basis matrices, considered as 2×2 complex matrices, shows that conjugate transposition merely negates them. Since each squares to −1, they — along with the identity — are all special unitary matrices, and Spin(3) is isomorphic to SU(2).

Mayer-Vietoris sequence

 * It's not much fun trying to write this stuff in wiki syntax, so I hesitate to dive in. Ah well, let's see what we can manage…
 * Let U be the left half of the torus plus a bit more, and V the right half plus a bit more. Both are open cylinders retracting to circles, and the same is true of their intersection, A, only doubled. For the Mayer-Vietoris sequence we'll need the two injections i: A→U and j: A→V, giving maps of cycle classes φn = (i∗,−j∗). The homology of S1 follows the pattern of spheres: H0 = Z, with generator cycle any point; and H1 = Z, with generator cycle the whole circle. For A, the pair of circles, we double these: both groups are Z+Z. All other homology groups of these three spaces are zero.
 * In particular, H2(U)⨁H2(V) is zero, so that makes a convenient place to begin the sequence.
 * {| style="text-align:center"


 * 0 ||   || ? || ∂ || Z+Z || φ || Z+Z ||   || ? || ∂ || Z+Z || φ || Z+Z ||   || Z
 * H2(U)⨁H2(V) || ⟶ || H2(X) || ⟶ || H1(A) || ⟶ || H1(U)⨁H1(V) || ⟶ || H1(X) || ⟶ || H0(A) || ⟶ || H0(U)⨁H0(V) || ⟶ || H0(X)
 * }
 * H2(U)⨁H2(V) || ⟶ || H2(X) || ⟶ || H1(A) || ⟶ || H1(U)⨁H1(V) || ⟶ || H1(X) || ⟶ || H0(A) || ⟶ || H0(U)⨁H0(V) || ⟶ || H0(X)
 * }


 * If you're not required to use Mayer-Vietoris, the two-triangle idea is temptingly straightforward. Take a rectangle with sides t, r, b, ℓ, and split it with a diagonal, d. We now have a pair of 2-simplexes (or 2-simplices, if you prefer), say t, r, d and b, ℓ, d. Glue t (top) to b (bottom), both oriented left-to-right; and glue r (right) to ℓ (left), both oriented top-to-bottom, to make this a torus.
 * Even using Mayer-Vietoris, notice that removing a disc is the same as deleting one of the triangles, say the left one. The glue must be retained; thus the top edge becomes a circle, and so does the right edge. The diagonal retracts so that all that remains is two circles; they share the point at the top right corner of the rectangle. Thus the homology of this part is that of a figure-8, S1∨S1, where "∨" denotes wedge sum. (This is exactly what EdC first suggested, just made slightly more explicit.)
 * The disc U is contractible, so has the homology of a point, while the intersection A has the homology of a circle, S1. Starting the sequence at H2 of U and V, we have
 * {| style="text-align:center"


 * 0 ||   || Z || ∂ || Z || φ || Z+Z ||   || Z+Z || ∂ || Z || φ || Z ||   || Z
 * H2(U)⨁H2(V) || ⟶ || H2(X) || ⟶ || H1(A) || ⟶ || H1(U)⨁H1(V) || ⟶ || H1(X) || ⟶ || H0(A) || ⟶ || H0(U)⨁H0(V) || ⟶ || H0(X)
 * }
 * H2(U)⨁H2(V) || ⟶ || H2(X) || ⟶ || H1(A) || ⟶ || H1(U)⨁H1(V) || ⟶ || H1(X) || ⟶ || H0(A) || ⟶ || H0(U)⨁H0(V) || ⟶ || H0(X)
 * }

Double sup
Let's try the suggested workaround for the Wikipedia bug involving nested sup/sub. Consider the function $x^{2}⁄2$. Now consider it without the  hack: $x^{2}⁄2$. Yippee!