The geometric algebra product of two vectors $v$ and $w$
$ v w = v \cdot w + v \wedge w $ is defined to be $ vw = 1/2 (vw +wv) + 1/2 ( vw- wv) $.
The Wedge product in differential geometry is defined as follows $ \vec{v} \wedge \vec{w} = \vec{v} \otimes \vec{w} - \vec{w} \otimes \vec{v} $
Except for the factor $1/2$, which could be a definition matter, the wedge product of two vectors in geometry algebra language and the wegde product in differential geometry language seem to be the same. Can I relate the GA product with the tensor product ? Can I interpret the inner product as a symmetric tensor product $ \vec{v} \cdot \vec{w} = \vec{v} \otimes \vec{w} + \vec{w} \otimes \vec{v} $ It seems like I would need $ \vec{v} \cdot \vec{w} = 1/2\text{Tr}(\vec{v} \otimes \vec{w} + \vec{w} \otimes \vec{v}) $.
Does anyone knows how this works or have a good source for me to learn ?
Geometric algebra (GA) is really just a Clifford algebra, i.e. it is basically a tensor algebra with an equivalence relation.
The product of two vectors $v$ and $w,$ written $vw,$ is defined as $v\otimes w$ with the extra condition that $v^2 := vv = |v|^2.$
The inner product (dot product) of two vectors is defined as $$v\cdot w = \frac12(vw+wv) = \frac12(v\otimes w + w\otimes v),$$ and the exterior product as $$v\wedge w = \frac12(vw-wv) = \frac12(v\otimes w - w\otimes v).$$
The exterior algebra of differential geometry is closely related. It is spanned by all antisymmetric elements in the tensor algebra, like $v\wedge w$ above and $$ u\wedge v\wedge w = \frac16(u\otimes v\otimes w - u\otimes w\otimes v + w\otimes u\otimes v - w \otimes v\otimes u + v\otimes w\otimes u - v\otimes u\otimes w). $$
The linked pages on Wikipedia have quite a lot of information.