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12 votes

Linearity of differential forms

The mapping $\omega:X=\Bbb{R}^3\to V^*\wedge V^*$ is not required to be linear, but for each point $x\in X$ the element $\omega_x$ belongs to $V^*\wedge V^*$; this space is canonically isomorphic to $(...
peek-a-boo's user avatar
6 votes
Accepted

Derivatives of the determinant of a singular matrix w.r.t the matrix

$ \def\o{{\tt1}} \def\t{{\large\tau}} \def\a{\t_1} \def\b{\t_2} \def\c{\t_3} \def\k{\t_k} \def\BR#1{\Big[#1\Big]} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\vc#1{\op{vec}\LR{#1}} \...
greg's user avatar
  • 36.8k
3 votes
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Proving a proposition on Lie derivatives of differential forms.

Cartan's magic formula says that for any differential form $\alpha$ and any vector field $X$, one has $$ L_X\alpha = d(\iota_X\alpha) + \iota_X(d\alpha). $$ In particular, if $\alpha = \omega\wedge\...
Didier's user avatar
  • 19.5k
2 votes
Accepted

Apparent dimensional problem when calculating a tensor contraction with matrices

The Wikipedia article is slightly confusing by omitting some brackets from the final tensor. It looks like a $4\times4$ matrix, i.e., a rank $2$ tensor with $4^2=16$ entries, but actually it is a $2\...
Lieven's user avatar
  • 1,953
1 vote
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Constructing a basis for a tensor product

If the pair $(V\otimes W,\otimes)$ is a tensor product of vector spaces $V$ and $W$ with tensor product map $\otimes:V\times W\to V\otimes W$, then the pair satisfies the universal property: For any ...
blargoner's user avatar
  • 3,180
1 vote
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Double Trace of the tensor product of the metric tensor with vector fields.

The notation you are using partly is not really well defined. There are (at least) two possible definitions of what $tr[g\otimes X\otimes Y]$ should mean, namely either $g(X,\ )\otimes Y$ or $g(\ ,Y)\...
Andreas Cap's user avatar
1 vote
Accepted

Mechanics of a contraction on the Kronecker product matrix

First a remark on the side: the distinction between upper and lower indices does not mean much in this context where we are not dealing with a metric; only the order of the indices matters. The result ...
Lieven's user avatar
  • 1,953
1 vote
Accepted

Abstract index notation - can't understand identity $(dx^{\mu})_aT^b\partial_bv^a=T^b\partial_b[(dx^{\mu})_av^a]$

Unfortunately the version of that book in google stops at p. 66. So I start from some principles that are tersely explained, say, in Wikipedia. If $\boldsymbol{v}=v^\nu\partial_\nu$ is a vector field ...
Kurt G.'s user avatar
  • 15.3k
1 vote
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trace of a matrix squared:a formula

We use the metric tensor to calculate the trace of a $2$-tensor: If $A=[a_{ij}]$ is a $2$-tensor, then $$\text{tr}(A) = \sum g^{ij}a_{ij}.$$ What we are doing is using the metric tensor to raise one ...
Ted Shifrin's user avatar
1 vote

Geometric meaning of second Covariant Derivative

If you consider the "second covariant derivative" of a scalar function $f$ instead of a vector field $w$, you get the nice symmetry result that $\nabla _{u,v}^{2}f=\nabla _{v,u}^{2}f$, ...
ansatz's user avatar
  • 11

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