Length of a Clifford Poly-vector? The length of a vector is defined as:
$$
||\mathbf{v}||^2=\mathbf{v}\cdot \mathbf{v}
$$
In the case that $\mathbf{v}:=a\hat{x}+b\hat{y}$ is expressed in an orthogonal basis using $\hat{x}$ and $\hat{y}$ as the generators of a Clifford algebra $Cl_{0,2}(\mathbb{R})$, then $||\mathbf{v}||^2=a^2+b^2$
For a poly-vector (say $\mathbf{p}:=c+a\hat{x}$), one can also define an inner product. Then if one takes the inner product, one gets
$$
||\mathbf{p}||^2=\mathbf{p}\cdot \mathbf{p}=c^2+a^2
$$
However, I am skeptical of the inner product defines the length of the poly-vector primarily become the scalar $1$ and the 1-vector basis $\hat{x}$ are not orthogonal. Intuitively I would think the square of the geometric product of $\mathbf{p}$ with itself would be the length.
$$
||\mathbf{p}||=\sqrt{\mathbf{p}\mathbf{p}}
$$
In the case where the vectors are orthogonal k-vectors, the result is the same. For example
$$
\sqrt{\mathbf{v}\mathbf{v}}=\sqrt{(a\hat{x}+b\hat{y})(a\hat{x}+b\hat{y})}\\
=aa\hat{x}\hat{x}+2ab(\hat{x}\hat{y}+\hat{y}\hat{x})+ bb\hat{y}\hat{y}\\
=\sqrt{a^2+b^2}
$$
But, in the case where the poly-vector is not a k-vector, the definition differs:
$$
\sqrt{\mathbf{p}\mathbf{p}}=\sqrt{(c+a \hat{x})(c+a \hat{x})}\\
\sqrt{c^2+2ac\hat{x}+ aa\hat{x}\hat{x}}\\
\sqrt{c^2+a^2+2ac\hat{x}}
$$
Using this definition, we conclude that in the case of a poly-vector, a scalar length cannot be defined. Thus, define such a line as the inner product ought to erase some important geometric information about the length of the poly-vector.
Is this correct?
Is there a standard definition for the length of a poly-vector?
 A: Let us assume that the quadratic space (V,g) over which the Clifford algebra $\mathcal{C}\ell(V)$ is being constructed is real. Just to fix the notation, we shall denote the Clifford product by juxtaposition. One can then define
\begin{align}
N:\mathcal{C}&\ell(V)\rightarrow\mathbb{R}\\
&a\mapsto \langle \tilde{a}a\rangle_0 \end{align}
where $\langle\;\cdot\;\rangle_0$ is the projection onto the scalar part and $\widetilde{ab}=\widetilde{b}\widetilde{a}$ is the reversion anti-automorphism. In particular, for $\alpha\in\mathbb{R}$ and for a basis $\{e_i\}$ there holds $\widetilde{\alpha}=\alpha$ and $\widetilde{e_i}=e_i$. Then, you would have defined a (squared) seminorm in the Clifford algebra. If $g$ is positive definite, then $N$ is indeed a norm. Namely, in the case $V=\mathbb{R}^2$ with the canonical inner product, an arbitrary element $\mathcal{C}\ell(V)\ni a=\alpha+a_1e_1+a_2e_2+b_{12}e_{12}$ has
$$N(a)=\alpha^2+a_1^2+a_2^2+b_{12}^2,$$
which is just the canonical inner product over the $2^n$-dimensional vector space structure of $\mathcal{C}\ell(V)$. Notice that for every $v\in \mathbb{R}^2$, there holds $N(v)=g(v,v)$, so $N$ is compatible to the original metric.

Trying to define a norm not considering the scalar part, I can only think of defining
\begin{align}
N':\mathcal{C}&\ell(V)\rightarrow\mathbb{R}\\
&a\mapsto N(a-\langle a \rangle_0) \end{align}
In the $V=\mathbb{R}^2$ case, if you would even want to discard the bivector part then you can make $N'(a)= N(a-\langle a \rangle_{0,2})$. But then again,
$$N'(a)=g(\langle a \rangle_1,\langle a \rangle_1),$$
which is just the norm in the vector part of $a$. 
