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Revision difference : Clifford algebra |
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Version 4 |
| Let $V$ be a vector space over a field $k$, and $Q:V\times V\to k$ a symmetric bilinear form. |
Let $V$ be a vector space over a field $k$, and $Q:V\times V\to k$ a symmetric bilinear form. |
| Then the Clifford algebra $\Cliff(Q,V)$ is the quotient of the tensor algebra $\mc{T}(V)$ by the |
Then the Clifford algebra $\Cliff(Q,V)$ is the quotient of the tensor algebra $\mc{T}(V)$ by the |
| relations |
relations |
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| $$v\otimes w+w\otimes v=-2Q(v,w)\qquad \forall v,w\in V.$$ |
$$v\otimes w+w\otimes v=-2Q(v,w)\qquad \forall v,w\in V.$$ |
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| Since the above relationship is not homogeneous in the usual $\Z$-grading on $\mc{T}(V)$, $\Cliff(Q,V)$ does not inherit a $\Z$-grading. However, by reducing mod 2, we also have a $\Z_2$-grading on $\mc{T}(V)$, and the relations above are homogeneous with respect to this, so $\Cliff(Q,V)$ has a natural $\Z_2$-grading, which makes it into a superalgebra. |
Since the above relationship is not homogeneous in the usual $\Z$-grading on $\mc{T}(V)$, $\Cliff(Q,V)$ does not inherit a $\Z$-grading. However, by reducing mod 2, we also have a $\Z_2$-grading on $\mc{T}(V)$, and the relations above are homogeneous with respect to this, so $\Cliff(Q,V)$ has a natural $\Z_2$-grading, which makes it into a superalgebra. |
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| In addition, we do have a filtration on $\Cliff(Q,V)$ (making it a filtered algebra), and the associated graded algebra $\Gr\Cliff(Q,V)$ is simply $\wedge^{\cdot}V$, the exterior algebra of $V$. In particular, $$\dim\Cliff(Q,V)=\dim\wedge^{\cdot}V=2^{\dim V}.$$ |
In addition, we do have a filtration on $\Cliff(Q,V)$ (making it a filtered algebra), and the associated graded algebra $\Gr\Cliff(Q,V)$ is simply $\wedge^{\cdot}V$, the exterior algebra of $V$. In particular, $$\dim\Cliff(Q,V)=\dim\wedge^{\cdot}V=2^{\dim V}.$$ |
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| The most commonly used Clifford algebra is the case $V=\R^n$, and $Q$ is the standard inner |
The most commonly used Clifford algebra is the case $V=\R^n$, and $Q$ is the standard inner |
| product with orthonormal basis $e_1,\ldots,e_n$. In this case, |
product with orthonormal basis $e_1,\ldots,e_n$. In this case, |
| the algebra is generated by $e_1,\ldots,e_n$ and the identity of the algebra $1$, with the relations |
the algebra is generated by $e_1,\ldots,e_n$ and the identity of the algebra $1$, with the relations |
| \begin{align*} |
\begin{align*} |
| e_i^2&=-1\\ |
e_i^2&=-1\\ |
| e_ie_j&=-e_je_i \quad (i\neq j) |
e_ie_j&=-e_je_i \quad (i\neq j) |
| \end{align*} |
\end{align*} |
| Trivially, $\Cliff(\R^0)=\R$, and it can be seen from the relations above that $\Cliff(\R)\cong\C$, the complex numbers, and $\Cliff(\R^2)\cong\mathbb{H}$, the quaternions. |
Thus, $\Cliff(\R)\cong\C$, the complex numbers, and $\Cliff(\R^2)\cong\mathbb{H}$, the quaternions. |
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| On the other hand, for $V=\C^n$ we get the particularly \PMlinkescapetext{simple} answer of |
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| $$\Cliff(\C^{2k}) \cong \mathrm{M}_{2^k}(\C) \qquad \Cliff(\C^{2k+1}) = \mathrm{M}_{2^k}(\C) \oplus \mathbf{M}_{2^k}(\C).$$ |
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