User:Silly rabbit/Sandbox

Siℓℓy &#8475;aввι&tau; ❤


 * User:Silly rabbit/Sandbox/Spinors
 * User:Silly rabbit/Sandbox/connection
 * User:Silly rabbit/Sandbox/Cartan connection
 * User:Silly rabbit/Sandbox/Affine connection
 * User:Silly rabbit/Sandbox/Cartan affine connection
 * User:Silly rabbit/Sandbox/List of missing differential geometry topics
 * User:Silly rabbit/Sandbox/Differentiable manifold
 * User:Silly rabbit/Sandbox/Moving frame
 * User:Silly rabbit/Sandbox/Hypernumber_History
 * User:Silly rabbit/Sandbox/Atiyah-Singer index theorem
 * User:Silly rabbit/Sandbox/Torsion tensor
 * User:Silly rabbit/Sandbox/Connection form
 * User:Silly rabbit/Sandbox/Parallel transport
 * User:Silly rabbit/Sandbox/Exterior algebra
 * User:Silly rabbit/Sandbox/Hyperbolic space
 * User:Silly rabbit/Sandbox/Template:Scroll box
 * User:Silly rabbit/Sandbox/Template:Printable
 * User:Silly rabbit/Sandbox/Printfooter problem

Proof of (1)


 * $$\begin{matrix} V & \to & C(q) \\ \downarrow & \swarrow & \\ A && \end{matrix} $$


 * $${{a \times n = } \atop {\ }} {{\underbrace{a + \cdots + a}} \atop n}$$


 * Doesn't work, &Ropf;n &Ropf;ℝn &#8477;&#8476;&#8475;

Proof of (1)

Proof of (2) First notice that for even powers
 * $$(\vec{a} \cdot \vec{\sigma})(\vec{b} \cdot \vec{\sigma}) \,$$
 * $$ = a_i \sigma_i b_j \sigma_j \,$$
 * $$ = a_i b_j \sigma_i \sigma_j \,$$
 * $$ = a_i b_j \left( \delta_{ij} + i \varepsilon_{ijk} \sigma_k \right) \,$$
 * $$ = a_i b_j \delta_{ij} + i \sigma_k \varepsilon_{ijk} a_i b_j \,$$
 * $$ = \vec{a} \cdot \vec{b} + i \vec{\sigma} \cdot ( \vec{a} \times \vec{b} )\,$$
 * }
 * $$ = a_i b_j \left( \delta_{ij} + i \varepsilon_{ijk} \sigma_k \right) \,$$
 * $$ = a_i b_j \delta_{ij} + i \sigma_k \varepsilon_{ijk} a_i b_j \,$$
 * $$ = \vec{a} \cdot \vec{b} + i \vec{\sigma} \cdot ( \vec{a} \times \vec{b} )\,$$
 * }
 * $$ = \vec{a} \cdot \vec{b} + i \vec{\sigma} \cdot ( \vec{a} \times \vec{b} )\,$$
 * }
 * $$ = \vec{a} \cdot \vec{b} + i \vec{\sigma} \cdot ( \vec{a} \times \vec{b} )\,$$
 * }
 * $$(\hat{n} \cdot \vec{\sigma})^{2n} = I \,$$

but for odd powers
 * $$(\hat{n} \cdot \vec{\sigma})^{2n+1} = \hat{n} \cdot \vec{\sigma} \,$$

Combine these two facts with the knowledge of the relation of the exponential to sine and cosine:

Which, when we use $$x = a (\hat{n} \cdot \sigma) \,$$ gives us
 * $$e^{ix} \,$$
 * $$= \sum_{n=0}^\infty{\frac{i^n x^n}{n!}} \,$$
 * $$= \sum_{n=0}^\infty{\frac{(-1)^n x^{2n}}{2n!}} + i\sum_{n=0}^\infty{\frac{(-1)^n x^{2n+1}}{(2n+1)!}} \,$$
 * }
 * $$= \sum_{n=0}^\infty{\frac{(-1)^n x^{2n}}{2n!}} + i\sum_{n=0}^\infty{\frac{(-1)^n x^{2n+1}}{(2n+1)!}} \,$$
 * }
 * $$= \sum_{n=0}^\infty{\frac{(-1)^n (a\hat{n}\cdot \sigma)^{2n}}{2n!}} + i\sum_{n=0}^\infty{\frac{(-1)^n (a\hat{n}\cdot \sigma)^{2n+1}}{(2n+1)!}} \,$$
 * $$= \sum_{n=0}^\infty{\frac{(-1)^n a^{2n}}{2n!}} + i (\hat{n}\cdot \sigma) \sum_{n=0}^\infty{\frac{(-1)^n a^{2n+1}}{(2n+1)!}} \,$$

The sum on the left is cosine, and the sum on the right is sine so finally,
 * $$e^{i a(\hat{n} \cdot \vec{\sigma})} = \cos{a} + i (\hat{n} \cdot \vec{\sigma}) \sin{a} \,$$

 A ration of food. (Silly rabbit 15:17, 20 June 2007 (UTC))