I know the basics of geometric algebra and that the geometric product is the sum of the inner and outer products. I've also seen the tensor product described as an outer product, which makes sense. Tensors themselves are defined as members of a vector space that's constructed as the tensor product of two other vector spaces. That all leads to me think that the tensor product is just a special case of the geometric product, just like the dot product, cross product, wedge product, etc. However, when I did a web search about tensors in geometric algebra, I found multiple sources saying that geometric algebra can be embedded into tensor algebra, but not vice versa -- that all multivectors are tensors but not vice versa. But how can multivectors, which are just objects built from the geometric product, be a subset of tensors? Tensors themselves are objects built from the tensor product, a type of outer product, and outer products themselves are just half of the geometric product?

  • $\begingroup$ Tensors are usually defined as multilinear maps from $V^*\times \cdots \times V^*\times V\times \cdots \times V\rightarrow {\Bbb R}$. Here, $V$ is a vector space and $V^*$ is its dual. How are geometric products defined? $\endgroup$ Jan 16, 2020 at 6:20
  • $\begingroup$ Are you asking because you don't know or are you prompting me to think about the definition? $\endgroup$ Jan 16, 2020 at 17:06
  • $\begingroup$ I'm trying to find out if you understand the concepts you talk about. For instance, what's the relation between tensors as they're usually defined and tensor product of vector spaces? $\endgroup$ Jan 16, 2020 at 23:40
  • 2
    $\begingroup$ Take care with the term "outer product". In most of mathematics, this means exactly the tensor product, whereas in geometric algebra, it used as a term for what is called the "exterior product" in the broader mathematics. $\endgroup$
    – qman
    Apr 25, 2020 at 0:55

2 Answers 2


Tensor product is not a type of outer product. You get the outer product by antisymmetrization of the tensor product. There is a universal property of the tensor product (of two spaces, say): any bilinear operation taking two vectors (one from each space) into the underlying field can be "modeled" on the tensor product (you can google universal property tensor product for details). Since the geometric product is bilinear, it can be factored through the canonical mapping $V\times W\to V\otimes W$ (namely, it is the composition of this mapping with a linear operator $V\otimes W\to Z$, where $Z$ is any vector space, in your case the whole algebra.

Geometric algebra includes the exterior algebra (antisymmetric tensors) but not all the tensors. The main point of geometric (Clifford) algebra is that it is an associative algebra, where elements have an inverse, etc.

  • $\begingroup$ Thanks for explaining! $\endgroup$ Jan 16, 2020 at 17:07
  • $\begingroup$ You are right, that was not accurate, but it goes to a vector space , namely the whole Clifford Algebra. The universal property holds anyways. I corrected my answer. $\endgroup$
    – GReyes
    Jan 20, 2020 at 3:58

On the contrary, the geometric product can be viewed as a special case of the tensor product. The geometric product has the same rules as the tensor product, except for one extra: For a vector v: \begin{equation} v^2=|v|^2 \end{equation}


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