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Geometric Product

Identity	Comment
$ab = a \cdot b + a \wedge b$	Geometric product of two vectors $a$ and $b$
$a \cdot b = \frac{1}{2}\left( ab + ba\right)$	Inner product of two vectors as symmetric part of geometric product
$a \wedge b = \frac{1}{2}\left( ab - ba\right)$	Exterior product of two vectors as antisymmetric part of geometric product

Multivectors

Identity	Comment
$a_1 \wedge\dots\wedge a_r = \frac{1}{r!} \sum_{\mathcal{K}} (-1)^\epsilon a_{k_1} \dots a_{k_r}$	Exterior product of $r$ vectors. $\mathcal{K}$ is the set of permutations of $1 \dots r$ and $\epsilon$ is even for even perms and odd for odd
$a_1 \wedge a_2 \wedge a_3 = \frac{1}{6} \left( a_1a_2a_3+a_3a_1a_2+a_2a_3a_1-a_1a_3a_2-a_2a_1a_3-a_3a_2a_1\right)$	Example of the above for $r=3$. Note $a \wedge b = \frac{1}{2}\left( ab - ba\right)$ is example for $r=2$
$a \cdot A_r = a A_r - (-1)^rA_r a \qquad (*)$	Prove by repeated application of $ab = 2a\cdot b - ba$ to $a a_1\dots a_r$ for orthogonal ${a_i}$
$a \wedge A_r = a A_r + (-1)^rA_r a \qquad (**)$	Extends definition of $\wedge$ to multivectors
$aA_r = a\cdot A_r + a\wedge A_r$	Combine above two formulae. $a\cdot A_r$ has grade $r-1$, $a\wedge A_r$ has grade $r+1$
$A_r\cdot(B_s\cdot C_t) = (A_r\wedge B_s)\cdot C_t \quad r+s\leq t \qquad (\triangleleft)$	Prove by grade projection
$A_r\cdot(B_s\cdot C_t) = (A_r\cdot B_s)\cdot C_t \quad r+t\leq s \qquad (\triangleright)$	Ditto
$a\cdot (a_1 \wedge\dots\wedge a_r) = \sum_{k=1}^r (-1)^{k+1} a\cdot a_k ; a_1 \wedge\dots\wedge \check a_k \wedge\dots\wedge a_r$	The check on $\check a_k$ means "omit that term from the exterior product"

Vectors and Bivectors

Identity	Comment
$a\cdot(b\wedge c) = a\cdot b;c - a\cdot c;b$	Simplest example of above identity, $r=2$
$a\cdot(a_1\wedge a_2\wedge a_3) = a\cdot a_1;a_2\wedge a_3 - a\cdot a_2;a_1\wedge a_3 + a\cdot a_3;a_1\wedge a_2$	Next simplest example, $r=3$
$(a\wedge b)\cdot B = a\cdot (b\cdot B)$	Application of $(\triangleleft)$ in the case $r=s=1$ and $t=2$
$(a\cdot B) \cdot b = a\cdot(B\cdot b)$	Application of $(\triangleright)$ in the case $r=t=1$ and $s=2$
$(a\wedge b)\cdot(c\wedge d) = b\cdot c; a\cdot d - a\cdot c ; b\cdot d$	Expand into geometric products and use $ab = 2a\cdot b - ba$
$a\cdot(b\wedge B) = a\cdot b; B - b\wedge(a\cdot B)$	Expand into geometic products using $(*)$ and $(**)$

Commutator Product

Identity	Comment
$M \times N = \frac{1}{2}(MN-NM)$	Definition, for multivectors $M$ and $N$
$L\times(M\times N) + N\times(L\times M) + M\times(N\times L) = 0$	Jacobi identity for multivectors $L$, $M$ and $N$
$B\times a = \frac{1}{2}(Ba-aB) = B \cdot a$	Special case for bivector $B$ and vector $a$ using $(*)$
$B\times A_r = \langle B\times A_r\rangle_r$	Commutator product with bivector is grade-preserving for blade $A_r$ and hence multivector $M$
$BA_r = B\cdot A_r + B\times A_r + B\wedge A_r$	Therefore $\frac{1}{2}(BA_r + A_rB) = B\cdot A_r + B\wedge A_r$
$(a\wedge b)\times B = (a\cdot B)\wedge b + a\wedge (b\cdot B)$	Expand into geometic products using $(**)$ and def'n of commutator product

Pseudoscalar

Identity	Comment
$I = e_1 \wedge\dots\wedge e_n$	For orthonormal vectors ${e_i | i=1\dots n}$
$I^2 = \pm1$	Sign depends on metric but is typically -1 in physics applications
$IA_r = (-1)^r(n-1)A_rI$	For blade $A_r$ in $n$-dims. $I$ always commutes with even-grade multivectors
$a\cdot (A_rI) = a\wedge A_r, I$	Using the above commutation rule together with $(*)$

Linear Algebra

Identity	Comment
$b = \mathsf F(a) $	Linear function (whose components form a matrix)
$\mathsf F(\mu a + \nu b) = \mu\mathsf F(a) + \nu\mathsf F(b)$	Linearity for vectors $a$ and $b$ and scalars $\mu$ and $\nu$
$\mathsf F(a\wedge b)=\mathsf F(a) \wedge \mathsf F(b)$	Action on a bivector
$\mathsf F(\mu M + \nu N) = \mu \mathsf F(M) + \nu \mathsf F(N) \ \mathsf F(A_r) = \langle\mathsf F(A_r)\rangle_r$	Linearity and grade-preservation for multivectors $M$ and $N$, and blade $A_r$
$a\cdot\overline{\mathsf F}(b) = \mathsf F(a)\cdot b$	Definition of $\overline{\mathsf F}$, the adjoint (transpose) of $\mathsf F$
$\langle A,\overline{\mathsf F}(B)\rangle = \langle\mathsf F(A) , B \rangle \qquad (\bullet)$	Multivector version of above definition of adjoint
$A_r\cdot \overline{\mathsf F}(B_s) = \overline{\mathsf F}\left(\mathsf F(A_r)\cdot B_s\right) \qquad r\leq s \qquad $	Combine definition of adjoint with $(*)$, eg for $A_r = a$ and $B_s = b\wedge c$
$\mathsf F(A_r)\cdot B_s = \overline{\mathsf F}\left(A_r\cdot \overline{\mathsf F}(B_s)\right) \qquad r\geq s \qquad (\dagger)$	Combine definition of adjoint with $(*)$, eg for $A_r = a\wedge b$ and $B_s = c$
$R = nm = \exp(-\hat B\theta/2) = \cos(\theta/2) - \hat B\sin(\theta/2)$	Rotor $R$ from vectors $m$ and $n$ where $m\cdot n = \cos\theta$, $\hat B = \frac{m\wedge n}{\sin(\theta/2)}$ and $\hat B^2 = -1$
$\mathsf F(M) = RM\tilde R$	Multivector $M$ rotated by angle $\theta$ in $\hat B$ plane

Determinant and inverse

Identity	Comment
$\mathsf F(I) = \det(\mathsf F), I$	Definition of determinant, as volume scale factor for $\mathsf F$
$\det(\mathsf F\mathsf G),I = \det(\mathsf F)\det(\mathsf G), I$	Because $\det(\mathsf F\mathsf G),I = \mathsf F\mathsf G(I) = \det(\mathsf G), \mathsf F(I) = \det(\mathsf F)\det(\mathsf G), I$ using the above definition
$\det(\mathsf F) = \det(\overline{\mathsf F})$	Because $\det(\mathsf F) = \langle\mathsf F(I)I^{-1}\rangle = \langle I\overline{\mathsf F}(I^{-1})\rangle = \langle \overline{\mathsf F}(I^{-1})I\rangle = \det(\overline{\mathsf F})$ using $(\bullet)$
$\mathsf F^{-1}(M) = \frac{1}{\det(\mathsf F)} I \overline{\mathsf F}(I^{-1}M)$	Inverse of $\mathsf F$. Holds also when $\mathsf F \rightarrow \overline{\mathsf F}$. From definition of determinant, multiply by arbitrary multivector $N$: $$\det(\mathsf F)IN = \mathsf F(I)N = \mathsf F\left(I\overline{\mathsf F}(N)\right)$$ using $(\dagger)$ and that $\mathsf F(I)\cdot N = \mathsf F(I)N$ and $I\cdot \overline{\mathsf F}(N) = I\overline{\mathsf F}(N)$. Now let $M = IN$ and apply $\mathsf F^{-1}$ to both sides.

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