Why are “innermorphisms” not useful?

I commonly studied type of linear function in geometric algebra is the outermorphism. For reference, here's Wikipedia's definition:

Let $f$ be an $\Bbb R$-linear map from $V$ to $W$. The outermorphism of $f$ is the unique map $\underline{\mathsf{f}} : \Lambda(V) \to \Lambda(W)$ satisfying $$\underline{\mathsf{f}}(x) = f(x)\\ \underline{\mathsf{f}}(A \wedge B) = \underline{\mathsf{f}}(A) \wedge \underline{\mathsf{f}}(B)\\ \underline{\mathsf{f}}(A + B) = \underline{\mathsf{f}}(A) + \underline{\mathsf{f}}(B)\\ \underline{\mathsf{f}}(1) = 1$$ for all vectors $x$ and all multivectors $A$ and $B$, where $\Lambda(V)$ denotes the exterior algebra over $V$.

Why do we not also study linear mappings which preserve the inner product: "innermorphisms"? We could define it analogously. The only change in the definition above would be the second equation would read $$\underline{\mathsf{f}}(A \cdot B) = \underline{\mathsf{f}}(A) \cdot \underline{\mathsf{f}}(B)$$

There are at least $2$ examples of these functions: both reflections and rotations preserve the inner product (and have all of the other properties listed above).

Is it that these are the only ones? Or that the only functions which are innermorphisms are also the orthogonal functions and thus we are already studying them?

I'm just not sure why outermorphisms are so useful (and they are), but that analogous functions which preserve the inner product are apparently not.

• If I understand correctly, you're describing an innermorphism as an automorphism of an inner product space (and also is the elements of the orthogonal group) - so things which are definitely studied. – Milo Brandt Sep 26 '15 at 15:19
• $A.B$ is a real number. So what is $f(A.B)$ ? Of course you could extend $f$ so that $f(1)=1$ (similarly to the exterior algebra case). Then $f(A.B)=f(A).f(B)$ is simply $f(A).f(B)=A.B$. Those $f$ are the isometries. – Roland Sep 26 '15 at 15:40
• @Roland $A \cdot B$ is not necessarily a real number in geometric algebra. If $A$ is a $k$-blade and $B$ is a $j$-blade, then $A\cdot B$ is a $(j-k)$-blade. If they happen to be the same size (as in $k=j$), then $A\cdot B$ is a scalar, but that does not have to be the case. – got it--thanks Sep 26 '15 at 15:44
• I think the point of outermorphisms is not primarily to preserve the wedge product, but to extend the function's domain from vectors to all multivectors. Any multivector is a sum of wedge products of vectors, but not a sum of dot products of vectors. So the function of a high-grade multivector cannot be defined by preserving the dot product. For plain vectors, $a\cdot b$ is a scalar, so $f(1)=1$ implies (by linearity) that $f(a\cdot b)=a\cdot b$ . Combined with $f(a\cdot b)=f(a)\cdot f(b)$, this makes it clear that $f$ is an isometry. – mr_e_man May 20 '18 at 7:48
• Actually, I don't see that the outermorphism must be completely linear; $f(1)=1$ implies (by addition) that $f(n)=n$ if $n$ is an integer, but not that $f(c)=c$ for arbitrary scalars $c$. – mr_e_man May 20 '18 at 7:59

We do, but for historical reasons they are called (linear) isometries.

• I'm looking for functions in which $(1)$ $f(a\cdot b) = f(a)\cdot f(b)$ holds. As far as I know, isometries are functions where $(2)$ $f(a)\cdot f(b)=a\cdot b$ holds. Are you saying these two statements are equivalent or just that $(1) \implies (2)$? – got it--thanks Sep 26 '15 at 15:24
• I was thinking in this sense: books.google.co.uk/… (see the bottom of the page and top of the next one. – Chappers Sep 26 '15 at 15:35
• @gotit--thanks: what are the domain and range of $f$ in item (1) in your comment intended to be? Note that $a.b \in \mathbb{R}$. – Rob Arthan Sep 26 '15 at 15:40
• @RobArthan $A \cdot B$ is not necessarily a real number in geometric algebra. If $A$ is a $k$-blade and $B$ is a $j$-blade, then $A\cdot B$ is a $(j-k)$-blade. If they happen to be the same size (as in $k=j$), then $A\cdot B$ is a scalar, but that does not have to be the case. The domain and range of $f$ are both $\Bbb G^n$. – got it--thanks Sep 26 '15 at 15:46
• @gotit--thanks: sorry I didn't spot that you were working in geometric algebra. – Rob Arthan Sep 26 '15 at 15:50

Setup

I suppose $$V$$ and $$W$$ are finite dimensional real vector spaces for convenience. I will use $$F$$ in place of $$\underline{\mathsf{f}}$$. To extend the inner product of vectors to all multivectors, I will follow the convention of Geometric Algebra book by MSE's own Alan Macdonald, where $$\cdot$$ is used to denote the "left contraction" ( sometimes denoted with the symbol $$\rfloor$$ ).

Outermorphisms

Why defined?

As pointed out by mr_e_man in a comment, this yields a very natural extension of a linear map $$f:V\to W$$ that is defined on all multivectors in $$\Lambda(V)$$. It's natural because it preserves the induced exterior algebra, as mentioned on Wikipedia.

Completely linear?

In a different comment, mr_e_man raised the question of whether an outermorphism must be linear from $$\Lambda(V)\to\Lambda(W)$$. Since $$F$$ is additive. This makes it linear over $$\mathbb Q$$, but not $$\mathbb R$$ (see, for example, this related MSE question).

I'll have to think more about this, but it may be forced that $$F(a)=a+g(a)I$$ where $$g(a)$$ is a scalar function (which is $$0$$ on every rational) and $$I$$ is a pseudoscalar for the image of the original linear map $$f$$.

To avoid the weird cases, it suffices to explicitly assume that $$F$$ is linear on scalars, since wedge products with scalars covers everything else. (Alternatively, throw away the axiom of choice and work in a model of $$\mathsf{ZF}$$ where $$\mathbb R$$-linearity is forced.)

An alternative definition

The standard definition in the question post begins with a linear map $$f:V\to W$$ and then uses some properties to extend it to all of $$\Lambda(V)$$ (with a new codomain of $$\Lambda(W)$$).

However, we could derive $$f$$ from $$F$$, suitably defined. We could say that an outermorphism is a map $$F:\Lambda(V)\to\Lambda(W)$$ satisfying $$F(A \wedge B) = F(A) \wedge F(B)\\ F(A + B) = F(A) + F(B)\\ F(a) = a\text{ for scalars }a\\ F(\mathbf v)\in W\text{ for }\mathbf v\in V$$

(Actually, I'm pretty sure "$$F(a)=aF(1)$$ and $$F(1)\ne0$$" could be used to replace "$$F(a) = a$$", but that requires some algebra.)

"Innermorphisms"

Another meaning

As an aside, "innermorphism" is sometimes used to mean "inner automorphism" (as in this MSE answer or these review notes) or something related (as in the GAP documentation). Here I am following the lead of the question and examining a very different meaning by analogy with the outermorphisms above.

Definition

In analogy to the definition of outermorphism above, how about

An innermorphism is a map $$F:\Lambda(V)\to\Lambda(W)$$ satisfying $$F(A \cdot B) = F(A) \cdot F(B)\\ F(A + B) = F(A) + F(B)\\ F(a) = a\text{ for scalars }a\\ F(\mathbf v)\in W\text{ for }\mathbf v\in V$$ (Again, I suspect "$$F(a)=aF(1)$$ and $$F(1)\ne0$$" could be used to replace "$$F(a) = a$$", but I've only verified this for $$W$$ up to $$2$$D.)

The core linear map

If we restrict $$F$$ to $$V$$, we have a linear map $$f$$. But, unlike in the outermorphism case, it's not arbitrary since $$F(\mathbf v)\cdot F(\mathbf w)=F(\mathbf v\cdot \mathbf w)=\mathbf v\cdot \mathbf w$$. Therefore, $$F$$ is an orthogonal transformation on $$V$$ (see, e.g. these notes).

Extending to multivectors

In the general case, it seems that (unlike outermorphisms) innermorphisms are not defined by their core linear map on ($$1$$-)vectors.

For example, suppose $$W$$ is at least four dimensional, with orthonormal set $$\mathbf w_1,\ldots,\mathbf w_4$$. And suppose $$V$$ is at least two dimensional, with $$\mathbf e_1,\mathbf e_2$$ an orthonormal pair so that $$F(\mathbf e_i)=\mathbf w_i$$ for $$i=1,2$$.

Then $$F(\mathbf e_2)=F(\mathbf e_1\cdot(\mathbf e_{1}\wedge\mathbf e_2))=F(\mathbf e_1)\cdot F(\mathbf e_{1}\wedge\mathbf e_2)=\mathbf w_1\cdot F(\mathbf e_{1}\wedge\mathbf e_2)$$ and $$F(\mathbf e_1)=F(-\mathbf e_2\cdot(\mathbf e_{1}\wedge\mathbf e_2))=-F(\mathbf e_2)\cdot F(\mathbf e_{1}\wedge\mathbf e_2)=-\mathbf w_2\cdot F(\mathbf e_{1}\wedge\mathbf e_2)$$.

This does not allow $$F(\mathbf e_{1}\wedge\mathbf e_2)$$ to be, say, $$\mathbf w_1\wedge\mathbf w_2+\mathbf w_1\wedge\mathbf w_3$$, but seems to allow it to be things like $$7+\mathbf w_1\wedge\mathbf w_2+5(\mathbf w_3\wedge\mathbf w_4)$$. And if $$W$$ has more dimensions, then a term like $$\mathbf w_3\wedge\mathbf w_4\wedge \mathbf w_5$$ is possible, too.

Why not defined?

In summary, innermorphisms restrict the core linear map to be orthogonal, and do not do enough to define the map on arbitrary multivectors (which is the point of outermorphisms), so they probably don't need to be defined.

• "By consideration of grade, $F(a)$ must be of the form $b+cf(v)$." What about a higher-grade multivector that contains $f(v)$ as a factor? – mr_e_man May 13 at 3:31
• By using duality ($B\mapsto BI$) and a few identities, the innermorphism condition becomes $$F(A\wedge B)\cdot F(I)=\big(F(A)\wedge F(B)\big)\cdot F(I)$$ I haven't figured out the further implications of this. – mr_e_man May 13 at 4:01
• @mr_e_man, both good points. I'll edit now to remove most of that stuff that falsely assumes something wrong about the grade. I'll think more later, but the whole argument might fall apart in that it needn't be $\mathbb R$-linear in a term of $I$. – Mark S. May 13 at 8:45