A surface, parametrized by $U$ and $V$, has a tangent bivector given by the wedge product
$$I = \vec x_U\wedge\vec x_V$$
where subscripts represent partial derivatives. The First Fundamental Form coefficients are defined as usual, but the Second Fundamental Form coefficients are vectors perpendicular to the surface:
$$E=\vec x_U\cdot\vec x_U=\lVert\vec x_U\rVert^2$$ $$F=\vec x_U\cdot\vec x_V$$ $$G=\vec x_V\cdot\vec x_V$$ $$\lVert I\rVert^2 = -I^2 = EG-F^2$$
$$\vec L=\vec x_{UU}^\perp=(\vec x_{UU}\wedge I)I^{-1}$$ $$\vec M=\vec x_{UV}^\perp$$ $$\vec N=\vec x_{VV}^\perp$$
If the surface is embedded in 3D space, then $\vec L$, $\vec M$, and $\vec N$ are all parallel to each other. In higher dimensions, they can be independent.
In terms of the above quantities, the Gaussian and mean curvatures are
$$K=\frac{\vec L\cdot\vec N-\vec M\cdot\vec M}{\lVert I\rVert^2}$$ $$\vec H=\frac{G\vec L-2F\vec M+E\vec N}{2\lVert I\rVert^2}$$
which are both invariant with respect to re-parametrization. But I found another such invariant while deriving a different formula* for $K$, and it has the same units as $K$ (inverse area). This quadvector quantity $Q$ seems to be another type of curvature, which is always $0$ in 3D.
$$Q = I\frac{(\vec L\wedge\vec M)G+(\vec M\wedge\vec N)E+(\vec N\wedge\vec L)F}{\lVert I\rVert^4}$$
I also discovered that $Q$ is proportional to the area of an ellipse traced by the normal curvature vector $(\frac{d^2\vec x}{ds^2}=k\hat n=\vec k)$ of geodesics with varying direction passing through the given point on the surface. (This ellipse also has neat relationships with $\vec H$ and $K$ .)
Has $Q$ ever been studied before? Is there a general method for finding or classifying such invariants of embedded manifolds?
* $$i = \frac I{\lVert I\rVert}$$ $$i\frac{\langle i_Ui_V\rangle_2}{\lVert\langle\vec x_U\vec x_V\rangle_2\rVert}=I\frac{\langle i_Ui_V\rangle_2}{\lVert I\rVert^2}=K-Q$$
(The angle brackets denote grade-projection in Geometric Algebra, and inside the brackets is a product of partial derivatives.)
See Hestenes' Tutorial, page 10 (Differential Geometry). (He uses "$I$" for the unit tangent bivector, where I use "$i$". Also, his multiplication order is $a\cdot\partial I = (a\cdot\partial)I \neq a\cdot(\partial I)$, which confused me at first.)
The quantity $K-Q$ is simply the coefficient of the curvature for a given bivector:
where $a$ and $b$ are tangent vectors (so their wedge product is proportional to $I$). The intrinsic curvature is $-K(a\wedge b)$, and the extrinsic curvature is $Q(a\wedge b)$, both of which are bivectors.
The curvature, $C(a\wedge b) = \frac{S(a)S(b)-S(b)S(a)}{2}$, is defined in terms of the "shape" $S$, which is the derivative of the projection operator:
$$P(a) = (a\cdot I)I^{-1} = (a\cdot i)(-i) = \frac{ai-ia}{2}(-i)$$ $$\nabla = \vec x^U\frac{\partial}{\partial U} + \vec x^V\frac{\partial}{\partial V} = \big(\vec x_VI^{-1}\big)\frac{\partial}{\partial U} + \big(-\vec x_UI^{-1}\big)\frac{\partial}{\partial V}$$ $$S(a) = \nabla P(a) = \nabla\big((a\cdot i)(-i)\big)$$ $$= \vec x^U\big((a\cdot i_U)(-i) + (a\cdot i)(-i_U)\big) + \vec x^V\big((a\cdot i_V)(-i) + (a\cdot i)(-i_V)\big)$$
(The superscript $\vec x^U$ denotes the reciprocal basis vector.) This works for any vector $a$, tangent or not. The scalar part of $S(a)$ turns out to be $2\vec H\cdot a$; the mean curvature $\vec H$ is orthogonal to the surface, so if $a$ is tangent to the surface, this scalar part disappears. Also, for tangent $a$, the shape can be calculated as a directional derivative (I might prove this in a later update):
$$S(a^\parallel) = -i(a\cdot\nabla)i$$
(only differentiating the right $i$, not the left $i$). In particular, $S(\vec x_U) = -i\frac{\partial i}{\partial U} = -ii_U$. We can differentiate the equation $ii = -1$ to get $i_Ui + ii_U = 0$; two bivectors' geometric product is anticommutative if and only if it is another bivector (no scalar or quadvector parts), so this implies that $S(\vec x_U)$ is a bivector. Explicitly,
$$i_U = \frac{\partial}{\partial U}\frac{I}{\sqrt{-I^2}} = \frac{I_U}{\sqrt{-I^2}} + I\frac{(-1/2)}{\sqrt{-I^2}^3}\big(-I_UI-II_U\big)$$ $$= \frac{I_U}{\sqrt{-I^2}} + \frac{I}{\sqrt{-I^2}^3}\big(\frac{I_UI+II_U}{2}\big)$$ $$= \frac{-I^2I_U + \frac12\big(II_UI+I^2I_U\big)}{\sqrt{-I^2}^3}$$ $$= \frac{\frac12\big(-I^2I_U+II_UI\big)}{-I^2\sqrt{-I^2}}$$ $$= \frac{\frac12I\big(II_U-I_UI\big)}{I^2\sqrt{-I^2}}$$ $$= \frac{\frac12I^{-1}\big(II_U-I_UI\big)}{\sqrt{-I^2}}$$ $$= \frac{I^{-1}\langle II_U\rangle_2}{\lVert I\rVert}$$ $$ii_U = \frac{I}{\lVert I\rVert}\frac{I^{-1}\langle II_U\rangle_2}{\lVert I\rVert} = \frac{\langle II_U\rangle_2}{\lVert I\rVert^2}$$ $$S(\vec x_U) = -ii_U = \frac{\langle I_UI\rangle_2}{\lVert I\rVert^2}$$
We also have $I = \vec x_U\wedge\vec x_V$, so
$$I_U = \vec x_{UU}\wedge\vec x_V + \vec x_U\wedge\vec x_{VU}$$ $$= (\vec x_{UU}^\parallel+\vec x_{UU}^\perp)\wedge\vec x_V + \vec x_U\wedge(\vec x_{VU}^\parallel+\vec x_{VU}^\perp)$$ $$= (\vec L\vec x_V - \vec M\vec x_U) + (\vec x_{UU}^\parallel\wedge\vec x_V + \vec x_U\wedge\vec x_{VU}^\parallel)$$
The terms on the right are tangent bivectors; their product with $I$ is a scalar, so it disappears in the grade-projection:
$$\langle I_UI\rangle_2 = \langle(\vec L\vec x_V - \vec M\vec x_U)I\rangle_2 + 0$$ $$\frac{\langle I_UI\rangle_2}{\lVert I\rVert^2} = \langle(\vec L\vec x_V - \vec M\vec x_U)\Big(\frac{I}{-I^2}\Big)\rangle_2$$ $$= \langle(\vec L\vec x_V - \vec M\vec x_U)(-I^{-1})\rangle_2$$ $$= \langle-\vec L(\vec x_VI^{-1}) - \vec M(-\vec x_UI^{-1})\rangle_2$$ $$= \langle-\vec L\vec x^U - \vec M\vec x^V\rangle_2$$ $$= -\vec L\vec x^U - \vec M\vec x^V = \vec x^U\vec L + \vec x^V\vec M$$
(The vectors are orthogonal, so their product is already a bivector.) This gives the formulas
(The second equation comes from swapping $U$ and $V$.) The shape is a linear function, so $S(a)$ for any tangent vector $a$ can be calculated from these. Now, the curvature $C$:
$$C(\vec x_U\wedge\vec x_V) = \frac{S(\vec x_U)S(\vec x_V)-S(\vec x_V)S(\vec x_U)}{2}$$ $$= \frac{\Big(-\vec x^U\vec x^U\Big)(\vec L\vec M-\vec M\vec L)+\Big(-\vec x^U\vec x^V\Big)(\vec L\vec N-\vec M\vec M)+\Big(-\vec x^V\vec x^U\Big)(\vec M\vec M-\vec N\vec L)+\Big(-\vec x^V\vec x^V\Big)(\vec M\vec N-\vec N\vec M)}{2}$$ $$= \Big(-\vec x^U\cdot\vec x^U\Big)(\vec L\wedge\vec M)+\frac12\Big(-\vec x^U\vec x^V\Big)(\vec L\cdot\vec N+\vec L\wedge\vec N-\vec M\cdot\vec M)+\frac12\Big(-\vec x^V\vec x^U\Big)(\vec M\cdot\vec M-\vec N\cdot\vec L-\vec N\wedge\vec L)+\Big(-\vec x^V\cdot\vec x^V\Big)(\vec M\wedge\vec N)$$ $$= \frac{-G}{\lVert I\rVert^2}\vec L\wedge\vec M-\frac{\vec x^U\vec x^V-\vec x^V\vec x^U}{2}(\vec L\cdot\vec N-\vec M\cdot\vec M)-\frac{\vec x^U\vec x^V+\vec x^V\vec x^U}{2}(\vec L\wedge\vec N)+\frac{-E}{\lVert I\rVert^2}\vec M\wedge\vec N$$ $$= \frac{-G}{\lVert I\rVert^2}\vec L\wedge\vec M-\vec x^U\wedge\vec x^V(\vec L\cdot\vec N-\vec M\cdot\vec M)-\vec x^U\cdot\vec x^V(\vec L\wedge\vec N)+\frac{-E}{\lVert I\rVert^2}\vec M\wedge\vec N$$ $$= \frac{-G}{\lVert I\rVert^2}\vec L\wedge\vec M-\frac{I}{\lVert I\rVert^2}(\vec L\cdot\vec N-\vec M\cdot\vec M)-\frac{-F}{\lVert I\rVert^2}(\vec L\wedge\vec N)+\frac{-E}{\lVert I\rVert^2}\vec M\wedge\vec N$$ $$= -\frac{G}{\lVert I\rVert^2}\vec L\wedge\vec M-IK-\frac{F}{\lVert I\rVert^2}(\vec N\wedge\vec L)-\frac{E}{\lVert I\rVert^2}\vec M\wedge\vec N$$ $$= -IK-\frac{\vec L\wedge\vec MG+\vec M\wedge\vec NE+\vec N\wedge\vec LF}{\lVert I\rVert^2}$$ $$= -IK+I^2\frac{\vec L\wedge\vec MG+\vec M\wedge\vec NE+\vec N\wedge\vec LF}{\lVert I\rVert^4}$$ $$= -IK+IQ = -I(K-Q) = -(K-Q)I = -(K-Q)(\vec x_U\wedge\vec x_V)$$
This, along with linearity of $S$ and $C$, proves the claim at the beginning of this answer.