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Wedge Product of Coordinate Basis One-Forms:
Define the wedge product of two one-form coordinate basis 1-forms to be antisymmetric:
$dx^a \wedge dx^b = -dx^b \wedge dx^a$
Wedge product of the identical one-form coordinate basis element:
$dx^a \wedge dx^a = -dx^a \wedge dx^a$
Therefore $dx^a \wedge dx^a = 0$
Define the wedge product to be linear, such that for scalar multiplication of $\alpha \in \mathbb{R}$
$(\alpha dx^a) \wedge dx^b = \alpha (dx^a \wedge dx^b)$
Also define linearity of wedge product for addition and wedge of two coordinate basis 1-forms:
$(dx^a + dx^b) \wedge dx^c = dx^a \wedge dx^c + dx^b \wedge dx^c$
Map the k-forms of coordinate basis onto the outer products of coordinate basis with the following:
$dx^{i_1} \wedge ... \wedge dx^{i_k} = k! dx^{[i_1} \otimes ... \otimes dx^{i_k]}$
$= \delta^{i_1 ... i_k}_{j_1 ... j_k} dx^{j_1} \otimes ... \otimes dx^{j_k}$
$= k! \delta^{i_1 ... i_k}_{|j_1 ... j_k|} dx^{|j_1} \otimes ... \otimes dx^{j_k|}$
Then we can see the degenerate k=2 case.
Also we see associativity.
linear combinations of basis forms / k-forms
A k-form is a linear combination of the basis elements.
Notice that only the increasing indexes are summed.
This excludes all duplicate sums with the antisymmetric basis:
$a = \underset{i_1 < i_2 < ... < i_k}{\Sigma} a_{i_1 ... i_k} dx^{i_1} \wedge ... \wedge dx^{i_k}$
$= a_{|i_1 ... i_k|} dx^{i_1} \wedge ... \wedge dx^{i_k}$
$= a_{|I|} dx^{\wedge I}$
You can sum all indexes, just make sure to normalize by the number of duplicates:
$= \frac{1}{k!} a_I dx^{\wedge I}$
$= \frac{1}{k!} a_{i_1 ... i_k} dx^{i_1} \wedge ... \wedge dx^{i_k}$
$= a_{i_1 ... i_k} dx^{[i_1} \otimes ... \otimes dx^{i_k]}$
$= a_{[i_1 ... i_k]} dx^{i_1} \otimes ... \otimes dx^{i_k}$
$= a_{i_1 ... i_k} dx^{i_1} \otimes ... \otimes dx^{i_k}$
Notice that the indexes of a k-form are antisymmetric, and that the k-form is a (0 k) tensor.
For summing p distinct antisymmetric indexes:
$A_{|a_1 ... a_p|} \delta^{|a_1 ... a_p| b_1 ... b_q}_{c_1 ... ... c_{p+q}}$
$= \frac{1}{p!} A_{a_1 ... a_p} \delta^{a_1 ... a_p b_1 ... b_q}_{c_1 ... c_p c_{p+1} ... c_{p+q}}$
Associativity based on the generalized Kronecker delta definition:
For p-form a, q-form b, r-form c:
$[( a \wedge b ) \wedge {c} ]_{a_1 ... a_{p+q+r}}
dx^{a_1} \otimes ... \otimes dx^{a_{p+q+r}}$
$= \delta^{d_1 ... d_p u_1 ... u_q v_1 ... v_r}_{e_1 ... ... ... e_{p+q+r}}
a_{|d_1 ... d_p|}
b_{|u_1 ... u_q|}
c_{|v_1 ... v_r|}
dx^{e_1} \otimes ... \otimes dx^{e_{p+q+r}}
$
$= [a \wedge (b \wedge {c})]_{e_1 ... e_{p+q+r}}
dx^{e_1} \otimes ... \otimes dx^{e_{p+q+r}}
$
Wedge product antisymmetry for p-form (i.e. (0 p) antisymmetric tensor) a and q-form (i.e. (0 q) antisymmetric tensor) b:
$ a \wedge b $
$ = (\frac{1}{p!} a_{u_1 ... u_p} dx^{u_1} \wedge ... \wedge dx^{u_p})
\wedge (\frac{1}{q!} b_{v_1 ... v_q} dx^{v_1} \wedge ... \wedge dx^{v_q}) $
if any of $ u_1 ... u_p, v_q ... v_q $ are not unique then the result is zero (courtesy of $ dx^u \wedge dx^u = 0 $ ).
Otherwise:
$ = \frac{1}{p!q!} a_{u_1 ... u_p} b_{v_1 ... v_q} (dx^{u_1} \wedge ... \wedge dx^{u_p}) \wedge (dx^{v_1} \wedge ... \wedge dx^{v_q}) $
Rotate the first $ dx^{v_1} $ to the far left side using $ dx^u \wedge dx^v = -dx^v \wedge dx^u $. Doing so requires $p$ applications of the rule.
$ = (-1)^p \frac{1}{p!q!} a_{u_1 ... u_p} b_{v_1 ... v_q} dx^{v_1} \wedge (dx^{u_1} \wedge ... \wedge dx^{u_p}) \wedge dx^{v_2} \wedge ... \wedge dx^{v_q} $
Continue to do the same for the rest of the $dx^{v_k}$'s, doung so a number of $q$ times:
$ = (-1)^{pq} \frac{1}{p!q!} a_{u_1 ... u_p} b_{v_1 ... v_q} (dx^{v_1} \wedge ... \wedge dx^{v_q}) \wedge (dx^{u_1} \wedge ... \wedge dx^{u_p}) $
$ = (-1)^{pq} (\frac{1}{q!} b_{v_1 ... v_q} dx^{v_1} \wedge ... \wedge dx^{v_q})
\wedge (\frac{1}{p!} a_{u_1 ... u_p} dx^{u_1} \wedge ... \wedge dx^{u_p}) $
$ = (-1)^{pq} b \wedge a $
Notice that by this rule, for 0-form scalar a and 1-form w, this means:
$a w = a \wedge w = w \wedge a = w a$
This means the wedge product of scalars and k-forms can act interoperable with scalar multiplication of k-forms.
For one-forms, our basis is a 1-form $e^a$ and our scalar components $w_a$ are 0-forms, so this works for scalar multiplication too:
$w = w_a e^a = w_a \wedge e^a = e^a \wedge w_a = e^a w_a$
Check out this discussion.
For (0 k) tensors in general, have the wedge product antisymmetrize them / map them onto k forms:
$a \wedge b$
$= (\frac{1}{k!} a_{[i_1 ... i_k]} dx^{i_1} \wedge ... \wedge dx^{i_k}) \wedge b$
$= (a_{[i_1 ... i_k]} dx^{i_1} \otimes ... \otimes dx^{i_k}) \wedge b$
Notice that scalars are considered 0-forms.
Wedge product of p-form i.e. (0 p) antisymmetric tensor a, and q-form i.e. (0 q) antisymmetric tensor b:
$a \wedge b$
$= (a_{i_1 ... i_p} dx^{i_1} \otimes ... \otimes dx^{i_p})
\wedge (b_{j_1 ... j_q} dx^{j_1} \otimes ... \otimes dx^{j_q})$
$= a_{|i_1 ... i_p|} dx^{|i_1} \wedge ... \wedge dx^{i_p|}
\wedge b_{|j_1 ... j_q|} dx^{|j_1} \wedge ... \wedge dx^{j_q|}$
$= \frac{1}{p!} a_{i_1 ... i_p} dx^{i_1} \wedge ... \wedge dx^{i_p}
\wedge \frac{1}{q!} b_{j_1 ... j_q} dx^{j_1} \wedge ... \wedge dx^{j_q}$
$= \frac{1}{p!q!} a_{i_1 ... i_p} b_{j_1 ... j_q} dx^{i_1} \wedge ... \wedge dx^{i_p} \wedge dx^{j_1} \wedge ... \wedge dx^{j_q}$
$= \frac{(p+q)!}{p!q!} a_{|i_1 ... i_p} b_{j_1 ... j_q|} dx^{|i_1} \wedge ... \wedge dx^{i_p} \wedge dx^{j_1} \wedge ... \wedge dx^{j_q|}$
$= \frac{(p+q)!}{p!q!} a_{i_1 ... i_p} b_{j_1 ... j_q} dx^{[i_1} \otimes ... \otimes dx^{i_p} \otimes dx^{j_1} \otimes ... \otimes dx^{j_q]}$
$= \frac{(p+q)!}{p!q!} a_{[i_1 ... i_p} b_{j_1 ... j_q]} dx^{i_1} \otimes ... \otimes dx^{i_p} \otimes dx^{j_1} \otimes ... \otimes dx^{j_q}$
Can also be thought of as:
$=
a_{|i_1 ... i_p|} b_{|j_1 ... j_q|}
\delta^{|i_1 ... i_p| |j_1 ... j_q|}_{k_1 ... k_p k_{p+1} ... k_{p+q}}
dx^{k_1} \otimes ... \otimes dx^{k_{p+q}}
$
$=
\frac{1}{p!q!} a_{i_1 ... i_p} b_{j_1 ... j_q}
\delta^{i_1 ... i_p j_1 ... j_q}_{k_1 ... k_p k_{p+1} ... k_{p+q}}
dx^{k_1} \otimes ... \otimes dx^{k_{p+q}}
$
$=
\frac{(p+q)!}{p!q!} a_{[i_1 ... i_p} b_{j_1 ... j_q]}
dx^{i_1} \otimes ... \otimes dx^{i_p} \otimes dx^{j_1} \otimes ... \otimes dx^{j_q}
$
Wedge Product of Coordinate Basis of Vector-Valued Forms:
Let our vector coordinate basis $\partial_i$ be treated as vector-valued 0-forms.
Define the wedge product to be antisymmetric on vectors:
$\partial_i \wedge \partial_j = -\partial_j \wedge \partial_i$
Generalize the definition as we did with the wedge product on coordinate one-forms:
$\partial_{i_1} \wedge ... \wedge \partial_{i_k}
= \frac{1}{k!} \partial_{[i_1} \otimes ... \otimes \partial_{i_k]}$
Treat our vector basis $\partial_a$ as vector-valued 0-forms.
Then for scalar $\phi$ we have $\partial_a \wedge \phi = \phi \wedge dx^a$
Therefore our scalar component representation of our vectors $v = v^a \partial_a$ can also be represented using wedge products:
$v = v^a \wedge \partial_a = \partial_a \wedge v^a$
Like with forms, have our wedge product antisymmetrize our vector.
Wedge produt of (p 0) tensor u and (q 0) tensor v:
$u \wedge v$
$=
u^{|i_1 ... i_p|} v^{|j_1 ... j_q|}
\delta^{k_1 ... k_p k_{p+1} ... k_{p+q}}_{|i_1 ... i_p| |j_1 ... j_q|}
\partial_{k_1} \otimes ... \otimes \partial_{k_{p+q}}
$
$=
\frac{1}{p!q!} u^{i_1 ... i_p} v^{j_1 ... j_q}
\delta^{k_1 ... k_p k_{p+1} ... k_{p+q}}_{i_1 ... i_p j_1 ... j_q}
\partial_{k_1} \otimes ... \otimes \partial_{k_{p+q}}
$
$=
\frac{(p+q)!}{p!q!} u^{[i_1 ... i_p} v^{j_1 ... j_q]}
\partial_{i_1} \otimes ... \partial_{i_p} \otimes \partial_{j_1} \otimes ... \otimes \partial_{j_q}
$
So for (p 0) tensor wedge (q 0) tensor:
$a \wedge b = (-1)^{pq} b \wedge a$
... right?
Does that mean that (based on the generalized Kronecker delta definition of the wedge product for (p q) tensorsverify:
(p q) tensor a and (r s) tensor b:
$a \wedge b = (-1)^{pr + qs} b \wedge a$
Ok this is one mystery ... in literature I only ever see the wedge product (anti)symmetry definition for k-forms, or (0 k) tensors, or vectors, or (k 0) tensors, but never (p q) tensors.
And it seems most literature defines it assuming a vector is antisymmetric, i.e. scaled by $(-1)^p$ for p=1.
However this gets countered when we get to the exterior derivative of vectors, which seems to be symmetric, and treats vectors as vector-valued 0-forms, and then treats p=0, such that $(-1)^p = 1$ and is symmetric.
Where am I going wrong?
TODO in this worksheet I am using generalized Kronecker delta identities but I don't define them until worksheet #5.
#5 is split between half being the Levi-Civita permutation tensor and half being the Hodge dual.
Should I move the generalized Kronecker delta stuff earlier? Or move this stuff later?
Honestly there's even a use of generalized Kronecker delta in worksheet #1 on tensors, and on wedge product in there too.
For a set of n vectors $u_i$, the i'th equal to $u_i = (u_i)^j \partial_j$:
$u_1 \wedge ... \wedge u_n$
$= ((u_1)^{i_1} \partial_{i_1}) \wedge ... \wedge ((u_1)^{i_n} \partial_{i_n})$
$= (u_1)^{i_1} \cdot ... \cdot (u_1)^{i_n} \partial_{i_1} \wedge ... \wedge \partial_{i_n}$
$= (u_1)^{i_1} \cdot ... \cdot (u_1)^{i_n} \cdot n! \partial_{[i_1} \otimes ... \otimes \partial_{i_n]}$
$= (u_1)^{j_1} ... (u_n)^{j_n}
\delta^{i_1 ... i_n}_{j_1 ... j_n}
\partial_{i_1} \otimes ... \otimes \partial_{i_n}$
... using $\delta^{i_1 ... i_n}_{j_1 ... j_n} = \epsilon^{i_1 ... i_n} \epsilon_{j_1 ... j_n}$ ...
$= \epsilon^{i_1 ... i_n} \epsilon_{j_1 ... j_n} (u_1)^{j_1} ... (u_n)^{j_n}
\partial_{i_1} \otimes ... \otimes \partial_{i_n}$
... using $\epsilon_{j_1 ... j_n} (u_1)^{j_1} ... (u_n)^{j_n} = det((u_a)^b)$ ...
$= det((u_a)^b) \epsilon^{i_1 ... i_n} \partial_{i_1} \otimes ... \otimes \partial_{i_n}$
$= det((u_a)^b) \epsilon_{1 ... n} \epsilon^{i_1 ... i_n} \partial_{i_1} \otimes ... \otimes \partial_{i_n}$
$= det((u_a)^b) \delta^{i_1 ... i_n}_{1 ... n} \partial_{i_1} \otimes ... \otimes \partial_{i_n}$
$= det((u_a)^b) n! \delta^{[i_1}_1 \cdot ... \cdot \delta^{i_n]}_n \partial_{i_1} \otimes ... \otimes \partial_{i_n}$
$= det((u_a)^b) n! \partial_{[1} \otimes ... \otimes \partial_{n]}$
... using (abusing?) wedge product for vectors: ...
$= det((u_a)^b) \partial_1 \wedge ... \wedge \partial_n$
$= det((u_a)^b) \star 1$
TODO put this after the Hodge dual definition?
Wedge of vectors and forms on coordinate basis
By the rule of p-forms wedge q-forms $a \wedge b = (-1)^{pq} b \wedge a$
Or by the rule of p-vectors wedge q-vectors $a \wedge b = (-1)^{pq} \wedge b \wedge a$
What about a 1-form $dx^i$ wedge a vector (-valued 0-form) $\partial_j$
Either rule would suggest that $dx^i \wedge \partial_j = \partial_j \wedge dx^i$
So is it symmetric?
If it was symmetric then the ordering of form vs vector basis elements would be interchangeable:
$dx^{i_1} \wedge ... \wedge dx^{i_p} \wedge \partial_{j_1} \wedge ... \wedge \partial_{j_q}$
$= \partial_{j_1} \wedge ... \wedge \partial_{j_q} \wedge dx^{i_1} \wedge ... \wedge dx^{i_p} $
Define the wedge product on any (p q) tensor as an antisymmetrizing operation:
$T \wedge b$
$= ({T^{i_1 ... i_p}}_{j_1 ... j_q}
\partial_{i_1} \otimes ... \otimes \partial_{i_p}
\otimes dx^{j_1} \otimes ... \otimes dx^{j_q}) \wedge b$
$= (\frac{1}{p!q!} {T^{i_1 ... i_p}}_{j_1 ... j_q}
\delta^{k_1 ... k_p}_{i_1 ... i_p}
\delta^{j_1 ... j_q}_{l_1 ... l_q}
\partial_{k_1} \otimes ... \otimes \partial_{k_p}
\otimes dx^{l_1} \otimes ... \otimes dx^{l_q}
) \wedge b
$
$= (\frac{1}{p!q!} {T^{i_1 ... i_p}}_{j_1 ... j_q}
\partial_{i_1} \wedge ... \wedge \partial_{i_p}
\wedge dx^{j_1} \wedge ... \wedge dx^{j_q}
) \wedge b
$
$= ({T^{i_1 ... i_p}}_{j_1 ... j_q}
\partial_{[i_1} \otimes ... \otimes \partial_{i_p]}
\otimes dx^{[j_1} \otimes ... \otimes dx^{j_q]}
) \wedge b
$
$= ({T^{[i_1 ... i_p]}}_{[j_1 ... j_q]}
\partial_{i_1} \otimes ... \otimes \partial_{i_p}
\otimes dx^{j_1} \otimes ... \otimes dx^{j_q}
) \wedge b
$
$= \frac{1}{p!q!} {T^{i_1 ... i_p}}_{j_1 ... j_q}
\delta^{k_1 ... k_p}_{i_1 ... i_p}
\delta^{j_1 ... j_q}_{l_1 ... l_q}
\partial_{k_1} \otimes ... \otimes \partial_{k_p}
\otimes dx^{l_1} \otimes ... \otimes dx^{l_q}$
This answer makes me think that the symmetric for forms-wedge-vectors is the correct one.
Generalize the wedge product for (p q) tensor a and (r s) tensor b in a coordinate basis:
$a \wedge b$
$= (
{a^{i_1 ... i_p}}_{j_1 ... j_q}
\partial_{i_1} \otimes ... \otimes \partial_{i_p}
\otimes dx^{j_1} \otimes ... \otimes dx^{j_q}
) \wedge (
{b^{k_1 ... k_r}}_{l_1 ... l_s}
\partial_{k_1} \otimes ... \otimes \partial_{k_r}
\otimes dx^{l_1} \otimes ... \otimes dx^{l_s}
)
$
$= \frac{1}{p!q!r!s!} (
{a^{i_1 ... i_p}}_{j_1 ... j_q}
\partial_{i_1} \wedge ... \wedge \partial_{i_p}
\wedge dx^{j_1} \wedge ... \wedge dx^{j_q}
) \wedge (
{b^{k_1 ... k_r}}_{l_1 ... l_s}
\partial_{k_1} \wedge ... \wedge \partial_{k_r}
\wedge dx^{l_1} \wedge ... \wedge dx^{l_s}
)
$
$= \frac{1}{p!q!r!s!}
{a^{i_1 ... i_p}}_{j_1 ... j_q}
{b^{k_1 ... k_r}}_{l_1 ... l_s}
\partial_{i_1} \wedge ... \wedge \partial_{i_p}
\wedge dx^{j_1} \wedge ... \wedge dx^{j_q}
\wedge
\partial_{k_1} \wedge ... \wedge \partial_{k_r}
\wedge dx^{l_1} \wedge ... \wedge dx^{l_s}
$
But what do we do about $\partial \wedge dx$? Is it symmetric or antisymmetric?
What if we use the generalized Kronecker delta definition?
$= \frac{1}{p!q!r!s!} (
{a^{i_1 ... i_p}}_{j_1 ... j_q}
\delta_{i_1 ... i_p}^{a_1 ... a_p}
\delta^{j_1 ... j_q}_{b_1 ... b_q}
\partial_{a_1} \otimes ... \otimes \partial_{a_p}
\otimes dx^{b_1} \otimes ... \otimes dx^{b_q}
) \otimes (
{b^{k_1 ... k_r}}_{l_1 ... l_s}
\delta_{k_1 ... k_r}^{c_1 ... c_r}
\delta^{l_1 ... l_s}_{d_1 ... d_s}
\partial_{c_1} \otimes ... \otimes \partial_{c_r}
\otimes dx^{d_1} \otimes ... \otimes dx^{d_s}
)
$
...and now we have something we can clearly anticommute thanks to the generalized Kronecker delta summations:
$= \frac{1}{p!q!r!s!}
{a^{i_1 ... i_p}}_{j_1 ... j_q}
{b^{k_1 ... k_r}}_{l_1 ... l_s}
\delta_{i_1 ... i_p}^{a_1 ... a_p}
\delta^{j_1 ... j_q}_{b_1 ... b_q}
\delta_{k_1 ... k_r}^{c_1 ... c_r}
\delta^{l_1 ... l_s}_{d_1 ... d_s}
\partial_{a_1} \otimes ... \otimes \partial_{a_p}
\otimes dx^{b_1} \otimes ... \otimes dx^{b_q}
\otimes
\partial_{c_1} \otimes ... \otimes \partial_{c_r}
\otimes dx^{d_1} \otimes ... \otimes dx^{d_s}
$
Wedge Product in Non-Coordinate Basis:
Wedge product in terms of outer products:
$e^a \wedge e^b$
By expansion of the basis:
$= ({e^a}_i dx^i) \wedge ({e^b}_j dx^j)$
By wedge product linearity:
$= {e^a}_i {e^b}_j (dx^i \wedge dx^j)$
By wedge product definition in terms of outer product:
$= {e^a}_i {e^b}_j (dx^i \otimes dx^j - dx^j \otimes dx^i)$
By antisymmetry:
$= -(
{e^b}_j dx^j \otimes {e^a}_i dx^i
- {e^a}_i dx^i \otimes {e^b}_j dx^j
)$
$= -(e^b \otimes e^a - e^a \otimes e^b)$
$= -e^b \wedge e^a$
Wedge product antisymmetry in terms of outer products for k-form:
Now that I think of it, this is pretty obvious once the 2-form case is explained.
$e^{i_1} \wedge ... \wedge e^{i_k}$
By expansion of the basis:
$= ({e^{i_1}}_{a_1} dx^{a_1}) \wedge ... \wedge ({e^{i_k}}_{a_k} dx^{a_k})$
By wedge product linearity:
$= {e^{i_1}}_{a_1} \cdot ... \cdot {e^{i_k}}_{a_k} (dx^{a_1} \wedge ... \wedge dx^{a_k})$
By wedge product definition in terms of outer product:
$= k! {e^{i_1}}_{a_1} \cdot ... \cdot {e^{i_k}}_{a_k} (dx^{[a_1} \otimes ... \otimes dx^{a_k]})$
By antisymmetry of the permutation of indexes, we can revere the $dx^{a_i}$ indexes to get:
$= (-1)^k k! {e^{i_1}}_{a_1} \cdot ... \cdot {e^{i_k}}_{a_k} (dx^{[a_k} \otimes ... \otimes dx^{a_1]})$
By wedge product linearity:
$= (-1)^k k! ({e^{i_k}}_{a_k} dx^{[a_k} \otimes ... \otimes {e^{i_1}}_{a_1} dx^{a_1]})$
By basis definition:
$= (-1)^k k! (e^{[i_k} \otimes ... \otimes e^{i_1]})$
By wedge product definition in terms of outer product:
$= (-1)^k e^{i_k} \wedge ... \wedge e^{i_1}$
Wedge product antisymmetry on any one-form:
$a \wedge b$
$= \overset{n}{\underset{i=1}{\Sigma}} (a_i e^i) \wedge \overset{n}{\underset{j=1}{\Sigma}} (b_j e^j)$
by wedge product linearity:
$= \overset{n}{\underset{i=1}{\Sigma}} \overset{n}{\underset{j=1}{\Sigma}} ((a_i e^i) \wedge (b_j e^j))$
$= \overset{n}{\underset{i=1}{\Sigma}} \overset{n}{\underset{j=1}{\Sigma}} (a_i b_j (e^i \wedge e^j))$
$= -\overset{n}{\underset{i=1}{\Sigma}} \overset{n}{\underset{j=1}{\Sigma}} (a_i b_j (e^j \wedge e^i))$
$= -\overset{n}{\underset{i=1}{\Sigma}} \overset{n}{\underset{j=1}{\Sigma}} ((b_j e^j) \wedge (a_i e^i))$
$= -b \wedge a$
Wedge product of two identical one-forms:
$a \wedge a$
$= \overset{n}{\underset{i=1}{\Sigma}} (a_i e^i) \wedge \overset{n}{\underset{j=1}{\Sigma}} (a_j e^j)$
$= \overset{n}{\underset{i=1}{\Sigma}} \overset{n}{\underset{j=1}{\Sigma}} ((a_i e^i) \wedge (a_j e^j))$
$= \overset{n}{\underset{i=1}{\Sigma}} \overset{n}{\underset{j=1}{\Sigma}} (a_i a_j (e^i \wedge e^j))$
Notice that, for $i=j, e^i \wedge e^i = 0$, and for $i \ne j, e^i \wedge e^j = -e^j \wedge e^i$ are both summed, with identical coefficients $a_i a_j$,
Therefore we get:
$a \wedge a = 0$
Ex:
$(dx + dy) \wedge (dx + dy)$
$= dx \wedge dx + dx \wedge dy + dy \wedge dx + dy \wedge dy$
using $dx \wedge dx = 0$ and $dx \wedge dy = -dy \wedge dx$:
$= dx \wedge dy - dx \wedge dy$
$= 0$
Notice that the wedge product of 2-forms with themselves is not necessarily zero:
$(dw \wedge dx + dy \wedge dz) \wedge (dw \wedge dx + dy \wedge dz)$
$=
dw \wedge dx \wedge dw \wedge dx
+ dw \wedge dx \wedge dy \wedge dz
+ dy \wedge dz \wedge dw \wedge dx
+ dy \wedge dz \wedge dy \wedge dz
$
$=
2 dw \wedge dx \wedge dy \wedge dz
$
By linearity of wedge product on coordinate basis one-forms, we see the linearity of non-coordinate basis one-forms:
$(\alpha e^i) \wedge e^j = (\alpha {e^i}_a dx^a) \wedge ({e^j}_b dx^b)$
By coordinate form linearity:
$(\alpha e^i) \wedge e^j = \alpha (({e^i}_a dx^a) \wedge ({e^j}_b dx^b))$
$(\alpha e^i) \wedge e^j = \alpha (e^i \wedge e^j)$
$(e^i + e^j) \wedge e^k = ({e^i}_a dx^a + {e^j}_b dx^b) \wedge ({e^k}_c dx^c)$
$= (({e^i}_a dx^a) \wedge dx^c + ({e^j}_b dx^b) \wedge dx^c) {e^k}_c$
$= ((({e^i}_a dx^a) \wedge dx^c) {e^k}_c + (({e^j}_b dx^b) \wedge dx^c) {e^k}_c)$
$= ({e^i}_a dx^a) \wedge ({e^k}_c dx^c) + ({e^j}_b dx^b) \wedge ({e^k}_c dx^c)$
$= e^i \wedge e^k + e^j \wedge e^k$
By linearity of wedge product, our rules for coordinate basis apply to non-coordinate basis as well:
$a = a_{i_1 ... i_k} e^{i_1} \otimes ... \otimes e^{i_k}$
$= a_{[i_1 ... i_k]} e^{i_1} \otimes ... \otimes e^{i_k}$
$= a_{i_1 ... i_k} e^{[i_1} \otimes ... \otimes e^{i_k]}$
$= \frac{1}{k!} a_{i_1 ... i_k} e^{i_1} \wedge ... \wedge e^{i_k}$
$= \frac{1}{k!} a_I e^{\wedge I}$
$= \underset{i_1 < i_2 < ... < i_k}{\Sigma} a_{i_1 ... i_k} e^{i_1} \wedge ... \wedge e^{i_k}$
$= a_{|i_1 ... i_k|} e^{i_1} \wedge ... \wedge e^{i_k}$
$= a_{|I|} e^{\wedge I}$
By linearity, coordinate to non-coordinate basis rules apply:
$a \wedge b = (-1)^{pq} b \wedge a$
Since k-wedge 1-forms are antisymmetric, then k-wedge of 1-form non-coordinate basis is antisymmetric.
$a \wedge b = (-1)^{pq} b \wedge a \rightarrow
(e^{i_1} \wedge ... \wedge e^{i_p}) \wedge (e^{j_1} \wedge ... \wedge e^{j_q})
= (-1)^{pq}
(e^{j_1} \wedge ... \wedge e^{j_q})
\wedge
(e^{i_1} \wedge ... \wedge e^{i_p})
$
interior product and wedge product
Notice that, despite the $\frac{1}{k!}$ in front of it, the application of a form to a vector is still the same:
$x \lrcorner a = x \lrcorner \frac{1}{k!} a_{i_1 ... i_k} (e^{i_1} \wedge ... \wedge e^{i_k})$
$= a_{i_1 ... i_k} x \lrcorner (e^{[i_1} \otimes ... \otimes e^{i_k]})$
$= a_{[i_1 ... i_k]} x \lrcorner (e^{i_1} \otimes ... \otimes e^{i_k})$
$= x^{i_1} a_{[i_1 ... i_k]} e^{i_2} \otimes ... \otimes e^{i_k}$
$= x^{i_1} a_{i_1 [i_2 ... i_k]} e^{i_2} \otimes ... \otimes e^{i_k}$
$= x^{i_1} a_{i_1 ... i_k} e^{[i_2} \otimes ... \otimes e^{i_k]}$
$= \frac{1}{(k-1)!} x^{i_1} a_{i_1 ... i_k} e^{i_2} \wedge ... \wedge e^{i_k}$
Partial-application of k-form is a generalization of the interior product:
$x \lrcorner a = a(x)$ for and vector x and k-form a.
$= x \lrcorner \frac{1}{k!} a_{i_1 ... i_k} e^{i_1} \wedge ... \wedge e^{i_k}$
$= x \lrcorner a_{i_1 ... i_k} e^{[i_1} \otimes ... \otimes e^{i_k]}$
$= x \lrcorner a_{[i_1 ... i_k]} e^{i_1} \otimes ... \otimes e^{i_k}$
$= x^{i_1} a_{[i_1 i_2 ... i_k]} e^{i_2} \otimes ... \otimes e^{i_k}$
$= x^{i_1} a_{i_1 [i_2 ... i_k]} e^{i_2} \otimes ... \otimes e^{i_k}$
since a is a k-form, a's indexes are antisymmetric:
$= x^{i_1} a_{i_1 [i_2 ... i_k]} e^{i_2} \otimes ... \otimes e^{i_k}$
$= x^{i_1} a_{i_1 i_2 ... i_k} e^{[i_2} \otimes ... \otimes e^{i_k]}$
use the identity of outer product to wedge product:
$= \frac{1}{(k-1)!} x^{i_1} a_{i_1 i_2 ... i_k} e^{i_2} \wedge ... \wedge e^{i_k}$
Mind you these $\frac{1}{k!}$ products are scaled out in the application of the k-form to a vector thanks to the wedge product definition with regards to the outer product.
So for k-form w and vectors $x_i \in \{x_1 ...x_k\}$, $(x_1 \lrcorner w)(x_2, ..., x_k) = w(x_1, x_2, ..., x_k)$
Also for vectors x and y and k-form w:
$x \lrcorner y \lrcorner w = -y \lrcorner x \lrcorner w$
$x \lrcorner x \lrcorner w = 0$ by antisymmetry of k-form indexes.
For vector x, p-form b, and q-form c:
$x \lrcorner (b \wedge c)$
$= x \lrcorner (
b_{i_1 ... i_p} e^{i_1} \wedge ... \wedge e^{i_p}
\wedge c_{j_1 ... j_q} e^{j_1} \wedge ... \wedge e^{j_q}
)$
$= x \lrcorner (
b_{i_1 ... i_p} e^{i_1} \wedge ... \wedge e^{i_p}
\wedge c_{j_1 ... j_q} e^{j_1} \wedge ... \wedge e^{j_q}
)$
...
$= x \lrcorner (
b_{|i_1 ... i_p|} e^{|i_1} \wedge ... \wedge e^{i_p|}
\wedge c_{|j_1 ... j_q|} e^{|j_1} \wedge ... \wedge e^{j_q|}
)$
$= x \lrcorner (
b_{|i_1 ... i_p} c_{j_1 ... j_q|} e^{|i_1} \wedge ... \wedge e^{i_p} \wedge e^{j_1} \wedge ... \wedge e^{j_q|}
)$
...
$= x \lrcorner (
\frac{1}{(p+q)!} b_{i_1 ... i_p} c_{j_1 ... j_q} e^{[i_1} \otimes ... \otimes e^{i_p} \otimes e^{j_1} \otimes ... \otimes e^{j_q]}
)$
$= b_{i_1 ... i_p} c_{j_1 ... j_q} (\frac{1}{(p+q)!} e^{[i_1} \otimes ... \otimes e^{i_p} \otimes e^{j_1} \otimes ... \otimes e^{j_q]}) (x^a e_a)$
...
by our definition of $b \wedge c$ in component form:
$= x \lrcorner (
\frac{(p+q)!}{p!q!} b_{[i_1 ... i_p} c_{j_1 ... j_q]} e^{i_1} \otimes ... \otimes e^{i_p} \otimes e^{j_1} \otimes ... \otimes e^{j_q}
)$
$= \frac{(p+q)!}{p!q!} b_{[i_1 ... i_p} c_{j_1 ... j_q]} (e^{i_1} \otimes ... \otimes e^{i_p} \otimes e^{j_1} \otimes ... \otimes e^{j_q}) (x^a e_a)$
by interior product definition:
$= \frac{(p+q)!}{p!q!} x^{i_1} b_{[i_1 ... i_p} c_{j_1 ... j_q]} e^{i_2} \otimes ... \otimes e^{i_p} \otimes e^{j_1} \otimes ... \otimes e^{j_q}$
...TODO...
$= \frac{(p+q)!}{p!q!} x^{k} b_{k [i_2 ... i_p} c_{j_1 ... j_q]} e^{i_2} \otimes ... \otimes e^{i_p} \otimes e^{j_1} \otimes ... \otimes e^{j_q}
+ \frac{(p+q)!}{p!q!} b_{[i_1 ... i_p|} x^{k} c_{k | j_2 ... j_q]} e^{i_1} \otimes ... \otimes e^{i_p} \otimes e^{j_2} \otimes ... \otimes e^{j_q}
$
...
$= \frac{(p+q)!}{p!q!} x^{k} b_{k [i_2 ... i_p} c_{j_1 ... j_q]} e^{i_2} \otimes ... \otimes e^{i_p} \otimes e^{j_1} \otimes ... \otimes e^{j_q}
+ (-1)^p \frac{(p+q)!}{p!q!} b_{[i_1 ... i_p|} x^{k} c_{k | j_2 ... j_q]} e^{i_1} \otimes ... \otimes e^{i_p} \otimes e^{j_2} \otimes ... \otimes e^{j_q}
$
...TODO...
$= (x \lrcorner b) \wedge c + (-1)^p b \wedge (x \lrcorner c)$
TODO what about:
$(x \wedge y) \lrcorner w$
Wedge products of sums of 0-forms to n-forms:
For $a = a^0 + {a^1}_i e^i + \frac{1}{2} {a^2}_{i_1 i_2} e^{i_1} \wedge e^{i_2} + ... + \frac{1}{n!} {a^n}_{i_1 ... i_n} e^{i_1} \wedge ... \wedge e^{i_n}$
$a \wedge b$
$= (
a^0
+ {a^1}_i e^i
+ \frac{1}{2} {a^2}_{i_1 i_2} e^{i_1} \wedge e^{i_2}
+ ...
+ \frac{1}{n!} {a^n}_{i_1 ... i_n} e^{i_1} \wedge ... \wedge e^{i_n}
) \wedge (
b^0
+ {b^1}_j e^j
+ \frac{1}{2} {b^2}_{j_1 j_2} e^{j_1} \wedge e^{j_2}
+ ...
+ \frac{1}{n!} {b^n}_{j_1 ... j_n} e^{j_1} \wedge ... \wedge e^{j_n}
)
$
The resulting terms that aren't zeroed will be all those whose sum of forms on the a and b side add up to ≤ n.
$\begin{array}{c|c|c|c|c|c}
\wedge
& b^0
& {b^1}_{j_1} e^{j_1}
& \frac{1}{2} {b^2}_{j_1 j_2} e^{j_1} \wedge e^{j_2}
& ...
& \frac{1}{(n-1)!} {b^{n-1}}_{j_1 ... j_{n-1}} e^{j_1} \wedge ... \wedge e^{j_{n-1}}
& \frac{1}{n!} {b^n}_{j_1 ... j_n} e^{j_1} \wedge ... \wedge e^{j_n}
\\
\hline
a^0
& a^0 b^0
& a^0 {b^1}_{j_1} e^{j_1}
& \frac{1}{2} a^0 {b^2}_{j_1 j_2} e^{j_1} \wedge e^{j_2}
& ...
& \frac{1}{(n-1)!} a^0 {b^{n-1}}_{j_1 ... j_{n-1}} e^{j_1} \wedge ... \wedge e^{j_{n-1}}
& \frac{1}{n!} a^0 {b^n}_{j_1 ... j_n} e^{j_1} \wedge ... \wedge e^{j_n}
\\
\hline
{a^1}_{i_1} e^{i_1}
& {a^1}_{i_1} b^0 e^{i_1}
& {a^1}_{i_1} {b^1}_{j_1} e^{i_1} \wedge e^{j_1}
& \frac{1}{2} {a^1}_{i_1} {b^2}_{j_1 j_2} e^{i_1} \wedge e^{j_1} \wedge e^{j_2}
& ...
& \frac{1}{(n-1)!} {a^1}_{i_1} {b^{n-1}}_{j_1 ... j_{n-1}} e^{i_1} \wedge e^{j_1} \wedge ... \wedge e^{j_{n-1}}
& 0
\\
\hline
\frac{1}{2} {a^2}_{i_1 i_2} e^{i_1} \wedge e^{i_2}
& \frac{1}{2} {a^2}_{i_1 i_2} b^0 e^{i_1} \wedge e^{i_2}
& \frac{1}{2} {a^2}_{i_1 i_2} {b^1}_{j_1} e^{i_1} \wedge e^{i_2} \wedge e^{j_1}
& \frac{1}{2^2} {a^2}_{i_1 i_2} {b^2}_{j_1 j_2} e^{i_1} \wedge e^{i_2} \wedge e^{j_1} \wedge e^{j_2}
& ...
& 0
& 0
\\
\hline
... &
\\
\hline
\frac{1}{(n-1)!} {a^{n-1}}_{i_1 ... i_{n-1}} e^{i_1} \wedge ... \wedge e^{i_{n-1}}
& \frac{1}{(n-1)!} {a^{n-1}}_{i_1 ... i_{n-1}} b^0 e^{i_1} \wedge ... \wedge e^{i_{n-1}}
& \frac{1}{(n-1)!} {a^{n-1}}_{i_1 ... i_{n-1}} {b^1}_{e_{j_1}} e^{i_1} \wedge ... \wedge e^{i_{n-1}} \wedge e^{j_1}
& 0
& ...
& 0
& 0
\\
\hline
\frac{1}{n!} {a^n}_{i_1 ... i_n} e^{i_1} \wedge ... \wedge e^{i_n}
& \frac{1}{n!} {a^n}_{i_1 ... i_n} b^0 e^{i_1} \wedge ... \wedge e^{i_n}
& 0
& 0
& ...
& 0
& 0
\end{array}$
or, reindexing :
$\begin{array}{c|c|c|c|c|c}
\wedge
& b^0
& {b^1}_{j_1} e^{j_1}
& \frac{1}{2} {b^2}_{j_1 j_2} e^{j_1} \wedge e^{j_2}
& ...
& \frac{1}{(n-1)!} {b^{n-1}}_{j_1 ... j_{n-1}} e^{j_1} \wedge ... \wedge e^{j_{n-1}}
& \frac{1}{n!} {b^n}_{j_1 ... j_n} e^{j_1} \wedge ... \wedge e^{j_n}
\\
\hline
a^0
& a^0 b^0
& a^0 {b^1}_{i_1} e^{i_1}
& \frac{1}{2} a^0 {b^2}_{i_1 i_2} e^{i_1} \wedge e^{i_2}
& ...
& \frac{1}{(n-1)!} a^0 {b^{n-1}}_{i_1 ... i_{n-1}} e^{i_1} \wedge ... \wedge e^{i_{n-1}}
& \frac{1}{n!} a^0 {b^n}_{i_1 ... i_n} e^{i_1} \wedge ... \wedge e^{i_n}
\\
\hline
{a^1}_{i_1} e^{i_1}
& {a^1}_{i_1} b^0 e^{i_1}
& {a^1}_{i_1} {b^1}_{i_2} e^{i_1} \wedge e^{i_2}
& \frac{1}{2} {a^1}_{i_1} {b^2}_{i_2 i_3} e^{i_1} \wedge e^{i_2} \wedge e^{i_3}
& ...
& \frac{1}{(n-1)!} {a^1}_{i_1} {b^{n-1}}_{i_2 ... i_n} e^{i_1} \wedge ... \wedge e^{i_n}
& 0
\\
\hline
\frac{1}{2} {a^2}_{i_1 i_2} e^{i_1} \wedge e^{i_2}
& \frac{1}{2} {a^2}_{i_1 i_2} b^0 e^{i_1} \wedge e^{i_2}
& \frac{1}{2} {a^2}_{i_1 i_2} {b^1}_{i_3} e^{i_1} \wedge e^{i_2} \wedge e^{i_3}
& \frac{1}{2^2} {a^2}_{i_1 i_2} {b^2}_{i_3 i_4} e^{i_1} \wedge e^{i_2} \wedge e^{i_3} \wedge e^{i_4}
& ...
& 0
& 0
\\
\hline
... &
\\
\hline
\frac{1}{(n-1)!} {a^{n-1}}_{i_1 ... i_{n-1}} e^{i_1} \wedge ... \wedge e^{i_{n-1}}
& \frac{1}{(n-1)!} {a^{n-1}}_{i_1 ... i_{n-1}} b^0 e^{i_1} \wedge ... \wedge e^{i_{n-1}}
& \frac{1}{(n-1)!} {a^{n-1}}_{i_1 ... i_{n-1}} {b^1}_{e_{i_n}} e^{i_1} \wedge ... \wedge e^{i_n}
& 0
& ...
& 0
& 0
\\
\hline
\frac{1}{n!} {a^n}_{i_1 ... i_n} e^{i_1} \wedge ... \wedge e^{i_n}
& \frac{1}{n!} {a^n}_{i_1 ... i_n} b^0 e^{i_1} \wedge ... \wedge e^{i_n}
& 0
& 0
& ...
& 0
& 0
\end{array}$
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