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4h
comment Struggling with connection between Clifford Algebra (/GA) and their matrix generators
Moreover, because $\gamma_a \cdot \gamma_b = 0$ for $a \neq b$, $\gamma_a \wedge \gamma_b = \gamma_a \gamma_b$. This means $\gamma_0 \wedge \gamma_1 \wedge \gamma_2 \wedge \gamma_3 = \gamma_0 \gamma_1 \gamma_2 \gamma_3$.
16h
comment Question about curvature calculation method in Lee's *Riemannian Manifolds* book
No, the Ricci tensor would be viewed as acting on line-like subspaces--also known as common vectors. - The definition most like that of the Riemann I gave above would just be a linear operator, mapping vectors in $T_pM$ to vectors in $T_p M$.
16h
revised Question about curvature calculation method in Lee's *Riemannian Manifolds* book
added a factor of the inner product of B with itself, to coincide with the usual definition of K
16h
answered Question about curvature calculation method in Lee's *Riemannian Manifolds* book
2d
comment What is the anticommutator of the interior product and codifferential (adjoint of exterior derivative)?
$\nabla \wedge X$ is grade-2.
Jan
25
answered What is the anticommutator of the interior product and codifferential (adjoint of exterior derivative)?
Jan
25
comment How to reach Moore-Penrose pseudoinverse solution to minimize error function
Okay, $X$ is a matrix, $X^T X$ is $m\times m$, so $w$ has $m$ elements, and $t$ has $n$ elements?
Jan
25
comment How to reach Moore-Penrose pseudoinverse solution to minimize error function
Sorry, both $x$ and $w$ are vectors? Then isn't $x^T w$ a number?
Jan
25
comment Why is the trace of the Riemann curvature tensor useful?
That is merely my own interpretation of the notion that the Ricci tensor is zero outside a matter distribution, but the Weyl tensor need not be. One could compare with electromagnetism problems: the Weyl tensor would be compared to the divergence- and curl-free part of the electric field, which is contributed by some exterior source, while the Ricci tensor is similar to the divergence-full part that comes from the charge density in the region of interest.
Jan
24
answered Why is the trace of the Riemann curvature tensor useful?
Jan
24
comment Relationship between Levi-Civita symbol and Grassmann numbers?
Yeah, bivectors can be identified with 2-index tensors. Though in pure exterior algebra, I'm not sure you can actually write out how to take a bivector and two vectors and return a scalar (that isn't zero).
Jan
24
comment Relationship between Levi-Civita symbol and Grassmann numbers?
No, Grassmann numbers (and their nontrivial products) are basis elements of a vector space (the exterior algebra). They are not like ordinary scalars in any sense. Each $\theta_i$ is characterized by $n$ real components, where $n$ is the number of distinct $\theta_i$. How could they be scalars? Scalar multiplication commutes.
Jan
24
answered What exactly are pseudovectors and pseudoscalars? And where could I read about them?
Jan
23
answered Question concerning tensors
Jan
22
answered $\det(I+\epsilon V)=1+\operatorname{trace}(V)\epsilon+O(\epsilon^2)$
Jan
18
answered Why is a linear transformation a (1,1) tensor?
Jan
18
answered Proof for the curl of a curl of a vector field
Jan
17
answered Relationship between Levi-Civita symbol and Grassmann numbers?
Jan
15
comment Physical components of a third-order tensor
Could you elaborate on what is meant by "physical components"?
Jan
15
answered Tensors as Multilinear maps?