Blade Products and Angles Between Subspaces

Abstract

Principal angles are used to define an angle bivector of subspaces, which fully describes their relative inclination. Its exponential is related to the Clifford geometric product of blades, gives rotors connecting subspaces via minimal geodesics in Grassmannians, and decomposes giving Plücker coordinates, projection factors and angles with various subspaces. This leads to new geometric interpretations for this product and its properties, and to formulas relating other blade products (scalar, inner, outer, etc., including those of Grassmann algebra) to angles between subspaces. Contractions are linked to an asymmetric angle, while commutators and anticommutators involve hyperbolic functions of the angle bivector, shedding new light on their properties.

Appendices

A. Appendix: Blade products in Grassmann algebra

We present some results of Sect. 5.1 in terms of Grassmann algebra, to make them more accessible to researchers who might not be familiarized with Clifford algebra. We provide direct proofs requiring (mostly) only Sect. 2.

Here X can be a real or complex vector space, with inner product $\langle \cdot ,\cdot \rangle$ (Hermitian product in the complex case, with conjugate-linearity in the first argument). The complex case is important for applications, but requires some adjustments in Sect. 2, which are indicated in footnotes.

Formulas relating the inner product of same grade blades to some angle are well known in the real case [10, 22]. The following one also works in complex spaces and for distinct grades.

Theorem A.1

$\langle A, B \rangle = \Vert A\Vert \Vert B\Vert \cos {\hat{\Theta }}_{A,B}$ for blades $A,B\in \bigwedge X$.

Proof

If grades are equal, follows from Lemma 2.6 (as $\langle A,B \rangle ={\tilde{A}}*B$), (4) and Definition 2.16. Otherwise both sides vanish. $\square$

The exterior product satisfies a similar formula.

Theorem A.2

$\Vert A\wedge B\Vert =\Vert A\Vert \Vert B\Vert \cos \Theta ^\perp _{[A],[B]}$ for blades $A,B\in \bigwedge X$.

Proof

Let $P^\perp =P_{B^\perp }$. Then $\Vert A\wedge B\Vert = \Vert (P^\perp A)\wedge B\Vert = \Vert P^\perp A\Vert \Vert B\Vert$, and the result follows from Proposition 2.10i. $\square$

This result corresponds to (13f), and our comments about that formula (after Corollary 5.3) also apply here.

Contraction or interior product by a vector is widely used in Geometry and Physics, but as its generalization for multivectors is less known, we give a brief description here. The following contraction is related to that of Sect. 5.1 by $A\lrcorner B = {\tilde{A}}\rfloor B$. For more details, see [6, 32].

Definition A.3

The (left) contraction $A\lrcorner B$ of $A\in \bigwedge ^p X$ on $B \in \bigwedge ^q X$ is the unique element of $\bigwedge ^{q-p} X$ such that $\langle C, A \lrcorner B \rangle = \langle A\wedge C, B \rangle$ for all $C\in \bigwedge ^{q-p} X$.

If $p=q$ then $A \lrcorner B = \langle A, B \rangle$, so the contraction generalizes the inner product for distinct grades, but giving a $(q-p)$-vector instead of a scalar. For $p\ne q$ this product is asymmetric (in general, $A\lrcorner B \ne B\lrcorner A$), with $A\lrcorner B = 0$ if $p>q$. In the complex case it is conjugate-linear in A and linear in B.

Let $A\in \bigwedge ^p X$ and $B \in \bigwedge ^q X$ be nonzero blades, and $B = B_P\wedge B_\perp$ be a PO decomposition¹⁷ of B w.r.t. A.

Proposition A.4

$A \lrcorner B = \langle A, B_P \rangle \, B_\perp$.

Proof

As $B_\perp$ is completely orthogonal to A, for any $C\in \bigwedge ^{q-p} X$ we have $\langle A\wedge C, B_P\wedge B_\perp \rangle = \langle A, B_P \rangle \langle C, B_\perp \rangle = \langle C, \langle A, B_P \rangle B_\perp \rangle$. $\square$

So $A\lrcorner B$ performs an inner product of A with a subblade of B where it projects, leaving another subblade of B completely orthogonal to A.

Corollary A.5

$B_\perp = B_P\lrcorner B/\Vert B\Vert ^2$.

Theorem A.6

$A \lrcorner B = \Vert A\Vert \Vert B\Vert \cos \Theta _{A,B}\, B_\perp$.

Proof

Follows from Propositions A.1, A.4 and 3.26 if $p\le q$, otherwise $\Theta _{A,B}=\frac{\pi }{2}$ and both sides vanish. $\square$

Corollary A.7

$A \lrcorner B = \epsilon _{A,B} \Vert P_{B} A\Vert \Vert B\Vert \, B_\perp$.

Corollary A.8

.

B. Appendix: Hyperbolic Functions of Multivectors

Hyperbolic functions of multivectors are defined as usual, in terms of exponentials or power series [16].

Definition B.1

For any $M\in \bigwedge X$, $\cosh M = \frac{e^M + e^{-M}}{2} = \sum \limits _{k=0}^\infty \frac{M^{2k}}{(2k)!}$ and $\sinh M = \frac{e^M - e^{-M}}{2} = \sum \limits _{k=0}^\infty \frac{M^{2k+1}}{(2k+1)!}$.

Though we only need the bivector case for Sect. 5.2, we present properties of these funcions in more generality, as the literature on them is rather scarce (a few other properties, mostly for blades, can be found in [16, p. 77]).

Proposition B.2

Let $M,N\in \bigwedge X$.

• (i) $\cosh M + \sinh M = e^M$.

• (ii) $\cosh M - \sinh M = e^{-M}$.

• (iii) $\cosh (-M) = \cosh M$ and $\sinh (-M) = -\sinh M$.

• (iv) $(\cosh M)^\sim = \cosh {\tilde{M}}$ and $(\sinh M)^\sim = \sinh {\tilde{M}}$.

• (v) If $M\times N=0$ then $\cosh (M)\times \sinh (N) = 0$.

• (vi) $\cosh ^2 M - \sinh ^2 M = 1$.

Proof

(i–iv) Immediate. (v) If $M\times N=0$ then $e^M e^N = e^{M+N}$, and so $\cosh M \sinh N = \left( e^{M+N}-e^{M-N}+e^{N-M}-e^{-M-N}\right) /4 = \sinh N \cosh M$. (vi) $\cosh ^2 M = (e^{2M}+2+e^{-2M})/4$ and $\sinh ^2 M = (e^{2M}-2+e^{-2M})/4$. $\square$

Proposition B.3

Let $H\in \bigwedge ^p X$ be homogeneous of grade p, and $r=p\!\!\mod 4$.

• (i) $(\cosh H)^\sim = \cosh H$ and $(\sinh H)^\sim = (-1)^{\frac{p(p-1)}{2}}\sinh H$.

• (ii) $\cosh H \in \bigoplus \limits _{k\in \mathbb {N}} \bigwedge ^{4k}X$ and $\sinh H \in \bigoplus \limits _{k\in \mathbb {N}} \bigwedge ^{4k+r}X$.

• (iii) If $r\ne 0$ then $\cosh H * \sinh H = 0$.

• (iv) If $r = 0$ then ${\left\{ \begin{array}{ll} \Vert \cosh H\Vert ^2 + \Vert \sinh H\Vert ^2 = \frac{\Vert e^H\Vert ^2 + \Vert e^{-H}\Vert ^2}{2}, \\ \Vert \cosh H\Vert ^2 - \Vert \sinh H\Vert ^2 = 1, \\ \Vert \cosh H\Vert \ge 1. \end{array}\right. }$

• (v) If $r = 1$ then ${\left\{ \begin{array}{ll} \Vert \cosh H\Vert ^2 + \Vert \sinh H\Vert ^2 = \Vert e^H\Vert ^2, \\ \Vert \cosh H\Vert ^2 - \Vert \sinh H\Vert ^2 = 1, \\ \Vert \cosh H\Vert \ge 1 \text { and } \Vert e^H\Vert = \Vert e^{-H}\Vert \ge 1. \end{array}\right. }$

• (vi) If $r = 2$ or 3 then ${\left\{ \begin{array}{ll} \Vert \cosh H\Vert ^2 + \Vert \sinh H\Vert ^2 = 1, \\ \Vert \cosh H\Vert ^2 - \Vert \sinh H\Vert ^2 = \langle e^{2H} \rangle _0, \\ \Vert \cosh H\Vert \le 1, \Vert \sinh H\Vert \le 1 \text { and } \Vert e^H\Vert = 1. \end{array}\right. }$

Proof

(i) Follows from Proposition B.2 iii and iv, as ${\tilde{H}}=(-1)^{\frac{p(p-1)}{2}} H$. (ii) Follows from i, as components of $\cosh H$ are even, and those of $\sinh H$ have the parity of p. (iii) By ii, these functions have no components of same grade when $r\ne 0$. (iv) ${\tilde{H}}=H$, so $4\Vert \cosh H\Vert ^2 = (e^{{\tilde{H}}} + e^{-{\tilde{H}}}) * (e^H + e^{-H}) = \Vert e^H\Vert ^2 + 2 + \Vert e^{-H}\Vert ^2$ and $4\Vert \sinh H\Vert ^2 = \Vert e^H\Vert ^2 - 2 + \Vert e^{-H}\Vert ^2$. (v) Likewise, but by ii and Proposition B.2i we also have $\Vert \cosh H\Vert ^2 + \Vert \sinh H\Vert ^2 = \Vert e^H\Vert ^2$, which implies $\Vert e^H\Vert = \Vert e^{-H}\Vert$. (vi) Now ${\tilde{H}}=-H$, so that $4\Vert \cosh H\Vert ^2 = 2 + \langle e^{2H}+e^{-2H} \rangle _0 = 2 + 2\langle \cosh (2H) \rangle _0$ and $4\Vert \sinh H\Vert ^2 = 2 - 2\langle \cosh (2H) \rangle _0$, while ii and Proposition B.2i imply $\langle \cosh (2H) \rangle _0 = \langle e^{2H} \rangle _0$. Finally, $\Vert e^H\Vert ^2 = e^{{\tilde{H}}}*e^H = \langle e^{-H}e^H \rangle _0 = 1$. $\square$

Note that (ii) and Proposition B.2i restrict which grades $e^H$ can include, and when $r \ne 0$ its components are divided between $\cosh H$ and $\sinh H$ according to their grades.