Back
Index notation:
${T^{a_1 ... a_p}}_{b_1 ... b_q}$ implicitly describes $T = {T^{a_1 ... a_p}}_{b_1 ... b_q} e^{a_1} \otimes ... \otimes e^{a_p} \otimes e_{b_1} \otimes ... \otimes e_{b_p}$
Abstract index notation:
Indexes are not enumerations,
nor are they related to any particular basis,
but are used to determine what space the tensor resides in:
${T^{a_1 ... a_p}}_{b_1 ... b_q} \leftrightarrow T(\cdot, ..., \cdot) \in \underbrace{V \otimes ... \otimes V}_{\times p} \otimes \underbrace{V^* \otimes ... \otimes V^*}_{\times q}$
In multi-index notation:
$T = {T^A}_B {e_A}^B$
$v_{(a_1 ... a_p)} = \frac{1}{p!} \cdot $ sum of even permutations of $a_1 ... a_p$ plus sum of odd permutations
$v_{[a_1 ... a_p]} = \frac{1}{p!} \cdot $ sum of even permutations of $a_1 ... a_p$ minus sum of odd permutations
$v_{|a_1 ... a_p|} = $ sum of even minus sum of odd permutations, only summing indexes in increasing order
$v_{(a_1 ... | a_k ... a_m | ... a_p)}$, $v_{[a_1 ... | a_k ... a_m | ... a_p]} = $ a sum of permutations on variables $a_1 ... a_p$ excluding the variables $a_k$ through $a_m$
completely symmetric tensor:
$T_{a_1 ... a_p} = T_{(a_1 ... a_p)}$
alternating / completely antisymmetric tensor:
$T_{a_1 ... a_p} = T_{[a_1 ... a_p]}$
degree-2 antisymmetric tensors are tensors such that $T_{ab} = -T_{ba}$:
$T_{[ab]} = \frac{1}{2} (T_{ab} - T_{ba}) = \frac{1}{2}(T_{ab} + T_{ab}) = T_{ab}$
$T_{(ab)} = \frac{1}{2} (T_{ab} + T_{ba}) = \frac{1}{2}(T_{ab} - T_{ab}) = 0$
degree-2 symmetric tensors are tensors such that $T_{ab} = T_{ba}$:
$T_{[ab]} = \frac{1}{2} (T_{ab} - T_{ba}) = \frac{1}{2}(T_{ab} - T_{ab}) = 0$
$T_{(ab)} = \frac{1}{2} (T_{ab} + T_{ba}) = \frac{1}{2}(T_{ab} + T_{ab}) = T_{ab}$
any degree-2 tensor can be represented as its symmetric and antisymmetric components:
$T_{(ab)} + T_{[ab]} = \frac{1}{2} (T_{ab} + T_{ba}) + \frac{1}{2} (T_{ab} - T_{ba})$
$= \frac{1}{2}(T_{ab} + T_{ab})$
$= T_{ab}$
p-form applies to a p-vector:
$
e^{a_1} \otimes ... \otimes e^{a_p} ( e_{b_1}, ... , e_{b_p} )
= e^{a_1} \otimes ... \otimes e^{a_p} ( e_{b_1} \otimes ... \otimes e_{b_p} )
= e^{a_1} (e_{b_1}) \cdot ... \cdot e^{a_p} (e_{b_p})
= \delta^{a_1}_{b_1} \cdot ... \cdot \delta^{a_p}_{b_p}
= \overset{p}{\underset{i=1}{\Pi}} \delta^{a_i}_{b_i}
$
In multi-index notation:
$e^A(e_B) = \overset{p}{\underset{i=1}{\Pi}} \delta^{a_i}_{b_i}$
Wedge product defined for one-form basis (more on wedge product later):
$e^{a_1} \wedge ... \wedge e^{a_p} = p! e^{[a_1} \otimes ... \otimes e^{a_p]}$
I'm sure there is an official multi-index way to do this, but here's mine that I just came up with:
$e^{\wedge A} = p! e^{[A]}$
Maybe a more-official way would be $\underset{i \in I}{\wedge} e^i$, but by this point I might as well just write out the ellipses. I'm looking for notation even more concise.
$e^{a_1} \wedge ... \wedge e^{a_p} = p! e^{[a_1} \otimes ... \otimes e^{a_p]}$
$= p! \delta^{[a_1}_{b_1} \cdot ... \cdot \delta^{a_p]}_{b_p} e^{b_1} \otimes ... \otimes e^{b_p}$
$= \delta^{a_1 ... a_p}_{b_1 ... b_p} e^{b_1} \otimes ... \otimes e^{b_p}$
$= p! \delta^{a_1 ... a_p}_{|b_1 ... b_p|} e^{b_1} \otimes ... \otimes e^{b_p}$
Differential form applied to outer product of vectors:
$e^{a_1} \wedge ... \wedge e^{a_p} (e_{b_1} \otimes ... \otimes e_{b_p})
= p! e^{[a_1} \otimes ... \otimes e^{a_p]} (e_{b_1} \otimes ... \otimes e_{b_p})
= p! \delta^{[a_1}_{b_1} \cdot ... \cdot \delta^{a_p]}_{b_p}
= \delta^{a_1 ... a_p}_{b_1 ... b_p}
= \delta^{a_1 ... a_p}_{|b_1 ... b_p|}
$
In multi-index notation:
$e^{\wedge A} (e_B) = p! e^{[A]} (e_B) = \delta^A_B$
Comma and semicolon notation:
$T_{,a} = e_a(T)$
$T_{;a} = \nabla_{e_a} T = \nabla_a T$
Comma / semicolon derivatives aren't used so often anymore, probably because it adds the new index to the right hand side, whereas derivative symbols add the new index to the left hand side
... and derivatives like the exterior derivative add the new $e^a \wedge$ to the left hand side as well ... maybe for consistency or for other reasons, $\nabla$ applies $e_a \otimes$ to the left hand side as well.
interior product
left interior product partial-application of forms:
p-form applies to a q-vector, when p > q:
$
e_B \lrcorner e^A
= (e_{b_1} \otimes ... \otimes e_{b_q}) \lrcorner (e^{a_1} \otimes ... \otimes e^{a_p})
= e^{a_1} \otimes ... \otimes e^{a_p} ( e_{b_1} \otimes ... \otimes e_{b_q} )
= e^{a_1} \otimes ... \otimes e^{a_p} ( e_{b_1}, ... , e_{b_q} )
= e^{a_1} \otimes ... \otimes e^{a_q} \otimes e^{a_{q+1}} \otimes ... \otimes e^{a_p} ( e_{b_1}, ... , e_{b_q} )
= e^{a_{q+1}} \otimes ... \otimes e^{a_p} \cdot \overset{q}{\underset{i=1}{\Pi}} \delta^{a_i}_{b_i}
$
p-form applies to a q-form when p < q:
$
e_B \lrcorner e^A
= (e_{b_1} \otimes ... \otimes e_{b_q}) \lrcorner (e^{a_1} \otimes ... \otimes e^{a_p})
= e^{a_1} \otimes ... \otimes e^{a_p} ( e_{b_1} \otimes ... \otimes e_{b_q} )
= e^{a_1} \otimes ... \otimes e^{a_p} ( e_{b_1}, ... , e_{b_q} )
= e^{a_1} \otimes ... \otimes e^{a_p} ( e_{b_1}, ... , e_{b_p}, e_{b_{p+1}}, ..., e_{b_q} )
= e_{b_{p+1}} \otimes ... \otimes e_{b_q} \cdot \overset{p}{\underset{i=1}{\Pi}} \delta^{a_i}_{b_i}
$
In multi-index notation, I'm not sure how to represent unions of sets of indexes.
$e_B \lrcorner e^{A C} = e^{A C}(e_B) = e^A(e_B) \cdot e^C = \overset{q}{\underset{i=1}{\Pi}} \delta^{a_i}_{b_i} \cdot e^C$
$e_{B C} \lrcorner e^A = e^A(e_{B C}) = e^A(e_B) \cdot e_C = \overset{p}{\underset{i=1}{\Pi}} \delta^{a_i}_{b_i} \cdot e_C$
Interior product of a vector with a (0 k) tensor:
$x \lrcorner a$
$= a_{i_1 ... i_k} (e^{i_1} \otimes ... \otimes e^{i_k}) (x^j e_j)$
$= a_{i_1 ... i_k} e^{i_1} (x^j e_j) \cdot e^{i_2} \otimes ... \otimes e^{i_k})$
$= a_{i_1 ... i_k} x^j \delta^{i_1}_j \cdot e^{i_2} \otimes ... \otimes e^{i_k})$
$= x^{i_1} a_{i_1 ... i_k} e^{i_2} \otimes ... \otimes e^{i_k}$
how about outer of vectors on (0 k)-tensor?
$(x \otimes y) \lrcorner w = y \lrcorner x \lrcorner w$
how about (p 0) tensor on (0 q) tensor, for p ≤ q:
$x \lrcorner a$
$= x^{i_1 ... i_p} a_{i_1 ... i_p i_{p+1} ... i_q} e^{i_{p+1}} \otimes ... \otimes e^{i_q}$
Right interior product (not sure what the symbol is):
$e_B \llcorner e^{C A} = e^{C A}(..., e_B) = e^C \cdot e^A(e_B) = \overset{q}{\underset{i=1}{\Pi}} \delta^{a_i}_{b_i} \cdot e^C$
$e_{C B} \llcorner e^A = e^A(..., e_{C B}) = e_C \cdot e^A(e_B) = \overset{p}{\underset{i=1}{\Pi}} \delta^{a_i}_{b_i} \cdot e_C$
Back