The Hodge star operator for people not quite in a hurry

Santiago Quintero de los Ríos for homotopico.com

July 24, 2019

Suppose that you have a k-form ˛ on a manifold M of dimension n. But you’re a spoiled brat and don’t like k-forms but rather .n k/-forms, so you go to your dad and throw a tantrum, and your dad says shhh, k-forms are okay, see? k-forms and .n k/-forms are basically the same, you see, k.M / and n k.M / even have the same dimension, it’s basically the same thing sweetie stop crying everyone’s staring at us please, but you won’t have it because all your friends have .n k/-forms and keep crying, and your dad already spent a lot of money in your k-form and .n k/-forms are so expensive! What can your dad do now? He takes your dumb k-form to his workshop and comes out three hours later with a shiny .n k/-form, and hands it to you smiling but he’s regretting having children, nay, having you at this point. Oh wow dad, that’s perfect thank you so much you’re the best dad, how did yo do it? Well I told you, k-forms and .n k/-forms are not that diﬀerent, you just need a metric and some patience and you can turn one into the other.

1 Metric on the exterior algebra

Let V be a ﬁnite-dimensional vector space of dimension n, and g a Lorentzian metric on V , i.e. a symmetric, k non-degenerate bilinear map g V V R. We can extend g bilinearly to ƒ V for any k as W !

g.u1 u ; w1 w / det.Œg.ui ; wj //; ^ ^ k ^ ^ k D where u1; : : : ; u ; w1; : : : ; w V and Œg.ui ; wj / is a matrix whose .i; j /-th entry is g.ui ; wj /. For k k 2 example, g.u1 u2; w1 w2/ g.u1; w1/g.u2; w2/ g.u1; w2/g.u2; w1/: ^ ^ D Now let ˛; ˇ ƒkV . With respect to some basis u1; : : : ; un of V (not necessarily orthonormal), we can 2 write 1 1 k ˛ ˛ ::: u u D kŠ 1 k ^ ^ 1 1 k ˇ ˇ ::: u u ; D kŠ 1 k ^ ^ and thus we can ﬁnd (in Einstein’s notation), writing gij g.ui ; uj / for the components of the metric in D

1 this basis, 1 i j g.˛; ˇ/ ˛ ::: ˇ ::: det.Œg.u ; u // D .kŠ/2 1 k 1 k 1 X 1.1/ k .k/ ˛ ::: ˇ ::: sgn./g : : : g D .kŠ/2 1 k 1 k Sk 2 1 X 1.1/ k .k/ ˛ ::: ˇ ::: sgn./g : : : g D .kŠ/2 1 k 1 k Sk 2 1 X .1/:::.k/ sgn./˛ ˇ ::: D .kŠ/2 1 k Sk 2 1 X 1:::k ˛ ˇ ::: D .kŠ/2 1 k Sk 2 1 1:::k ˛ ˇ ::: : D kŠ 1 k Here we used the fact that the components of a form are totally antisymmetric, so for any permutation S 2 k

˛ ::: sgn./˛ ::: : .1/ .k/ D 1 k k k With this result we can see that g ƒ V ƒ V R is non-degenerate. Choose an orthonormal basis W ! e1; : : : ; en of V . Then we have X g.e1 ek ; e1 ek / sgn./g1.1/ : : : gk .k/ : ^ ^ ^ ^ D Sk 2 However, since the basis is orthonormal then gii 1 and gij 0 if i j . From this we see that if D ˙ D ¤ 1; : : : ; 1; : : : ; , then for absolutely no permutation S we will have 1 and ::: f kg ¤ f kg 2 k D .1/ and . Thus the inner product is nonzero only if 1; : : : ; 1; : : : ; . In this case, then, k D .k/ f kg D f kg we have that .1; : : : ; k/ is precisely a permutation of .1; : : : ; k/. Since all symbols 1; : : : ; k must be distinct (otherwise e1 ek is zero to begin with), we then have that there is only one permutation ^ ^ that survives in the sum, the one for which precisely i . In conclusion: .i/ D ( s . 1/ sgn./ if there exists such that j g.e1 ek ; e1 ek / .j / D ; ^ ^ ^ ^ D 0 otherwise where here s is the number of elements in e1 ; : : : ; ek which have negative length. Since we know that f g the elements of the form e1 ek form a basis for ƒkV , the previous result tells us that in this basis ^ ^ the matrix of g is diagonal with entries 1, and thus g is non-degenerate. ˙ 2 Deﬁning the Hodge star

n Now let vol ƒ V be a volume form on V , given in terms of an oriented orthonormal basis e1; : : : ; en as 2

vol e1 en: D ^ ^ We now deﬁne the Hodge star operator ? ƒkV ƒn kV , as the unique linear operator such that for all W ! ˛; ˇ ƒkV , 2 ˛ ?ˇ g.˛; ˇ/vol: ^ D

2 Here we’ve sneakily claimed that such a linear operator exists and is unique. We need to prove that. First, n k k for each ˇ ƒ V deﬁne a map ˇ ƒ V R such that 2 W ! ˛ ˇ .˛/vol: ^ D ˇ This map is well-deﬁned and clearly linear, i.e. .ƒkV/ . In particular, we can see that in components ˇ 2 with respect to an orthonormal basis e1; : : : ; en of V , 1 1:::k 1:::n k ˛ ˇ ˛1:::k ˇ1:::n k vol; ^ D kŠ.n k/Š where 1:::n is the Levi-Civita symbol, so that 1 1:::k 1:::n k ˇ .˛/ ˛1:::k ˇ1:::n k : D kŠ.n k/Š Now let’s see that the assignment, let’s call it ƒn kV .ƒkV/ , given as ˇ is an isomorphism. W ! 7! ˇ First, it is clearly linear. Now suppose that 0, i.e. for all ˛ ƒkV , .˛/ 0. In particular, for ˇ D 2 ˇ D ˛ e1 ek , D ^ ^ 1 1 k 1 k 1:::k 1:::n k 0 ˇ .e e / .e e /1:::k ˇ1:::n k : D ^ ^ D kŠ.n k/Š ^ ^ Now we have that the components of the basis itself are ( 0 if ; : : : ; ; : : : ; 1 k 1 k 1 k .e e /1:::k f g ¤ f g : ^ ^ D sgn./ if S such that i 2 k D .i/ We denote the right-hand side as the following symbol: ( 0 if ; : : : ; ; : : : ; 1:::k 1 k 1 k ı1:::k f g ¤ f g : WD sgn./ if S such that i 2 k D .i/ A little bit of tedious work shows that ı1:::k det.Œıi /; 1:::k D j i i.e. the determinant of the matrix whose .i; j /-th entry is ıj . Now when we plug this back in, we get 1 1 k 1:::k 1:::k 1:::n k 0 ˇ .e e / ı1:::k ˇ1:::n k : D ^ ^ D kŠ.n k/Š Here we are implicitly summing over all the i indices. From the deﬁnition above, the only terms that survive in the sum are those for which there exists a permutation S such that i . Therefore, 2 k D .i/ 1 1 k X .1/:::.k/1:::n k ˇ .e e / ˇ1:::n k sgn./ ^ ^ D kŠ.n k/Š Sk 2 1 X 1:::k 1:::n k ˇ1:::n k D kŠ.n k/Š Sk 2 1 1:::k 1:::n k ˇ1:::n k D .n k/Š For each set of indices 1; : : : ; n k if we choose 1; : : : ; k complementary to 1; : : : ; n k in f 1:::k1g:::n k f g f g 1; : : : ; n , then we have that 1 and so ˇ1:::n k 0. Therefore ˇ 0. The map f g D ˙ D D 3 ˇ is injective, then, and since dim.ƒn kV/ dim.ƒkV/ dim..ƒkV/ /, we obtain that it is W 7! ˇ D D an isomorphism. Recall that we have a metric g on ƒkV , which induces an isomorphism g ƒkV .ƒkV/ , given as [ W 7! g .˛/ g. ; ˛/. We deﬁne then ? ƒkV ƒn kV as ?˛ being the unique element in ƒn kV such that [ WD W ! ?˛ g .˛/: D [ If you want to, you could write ? 1 g . At once, this tells us that for any ˛; ˇ ƒkV , D ı [ 2 ˛ ?ˇ .˛/vol g .ˇ/.˛/vol g.˛; ˇ/vol: ^ D ?ˇ D [ D Okay so the map exists. What about uniqueness? Suppose there is an isomorphism ƒkV ƒn kV W ! such that ˛ .ˇ/ g.˛; ˇ/vol. This tells us that .˛/ g.˛; ˇ/, i.e. that g .ˇ/. But then, ^ D .ˇ/ D .ˇ/ D [ by our deﬁnition of ?, this precisely means that .ˇ/ ?ˇ. D Now a quick example which will help us down the road: We want to compute ?.e1 ek /. We ^ ^ use the fact that ? is an isomorphism, so we can make an educated guess and just check it works. Whatever it is, it has to satisfy

.e1 ek / ?.e1 ek / g.e1 ek ; e1 ek /vol . 1/svol; ^ ^ ^ ^ ^ D ^ ^ ^ ^ D where s is, again, the number of elements in e1 ; : : : ; ek with negative length, or . 1/s g11 : : : gk k . f g D This means that ?.e1 ek / has to consist of the wedges of the basis elements that we don’t have in 1 k ^ ^ e ; : : : ; e . That is, let 1; : : : ; n k be complementary to 1; : : : ; k in 1; : : : ; n . Therefore, f g f g f g e1 ek e1 en k 1:::k 1:::n k vol: ^ ^ ^ ^ ^ D With this, we then see that

?.e1 ek / . 1/s1:::k 1:::n k .e1 en k / (no Einstein sum): ^ ^ D ^ ^ 3 Making it useful: formulas in coordinates

This is all nice and all but we want to compute the star of a form explicitly if we have it in terms of some basis. Can do! Let e1; : : : ; en be an orthonormal basis of V . By deﬁnition, we have for any ˛; ˇ ƒkV , 2 that ˛ ?ˇ g.˛; ˇ/vol: ^ D In components with respect to the orthonormal basis, this is 1 1 1:::k 1:::n k 1:::k ˛1:::k .?ˇ/1:::n k ˛ ˇ1:::k : kŠ.n k/Š D kŠ Again, we choose ˛ e1 ek , so that we obtain D ^ ^ 1 1 1:::k 1:::k 1:::n k 11 k k 1:::k ı .?ˇ/ ::: g : : : g ı ˇ ::: : 1:::k 1 n k 1::: 1 k kŠ.n k/Š D kŠ k On the right-hand side, we have

11 k k 1:::k 1:::k 1:::k g : : : g ı ˇ ::: ı ˇ : 1:::k 1 k D 1:::k 1:::k But now recall that ı is non-zero only when there is a permutation such that i .i/. Then we 1:::k D have

::: X X X ı 1 k ˇ1:::k ı1:::k ˇ.1/:::.k/ sgn./ˇ.1/:::.k/ sgn./2ˇ1:::k kŠˇ1:::k : 1:::k D .1/:::.k/ D D D Sk Sk Sk 2 2 2 4 On the left-hand side, we have something similar:

ı1:::k 1:::k 1:::n k kŠ1:::k 1:::n k : 1:::k D When we put it all together, we get 1 1:::k 1:::n k 1:::k .?ˇ/1:::n k ˇ : .n k/Š D We are nearly done! We only need to get rid of that Levi-Civita symbol on the left-hand side. To do so, consider the sum1 1:::k 1:::n k 1:::k 1:::n k : The only terms of the sum that are non-zero are when 1; : : : ; is complementary to both the sets f kg 1; : : : ; n k and 1; : : : ; n k in 1; : : : ; n . This implies that the sum is non-zero only if 1; : : : ; n k f g f g f g f g D 1; : : : ; n k . Then suppose that there is some Sn k such that i .i/. Without the Einstein 2 D fconvention, thisg becomes:

X 1:::k 1:::n k X 1:::k 1:::n k 1:::k .1/:::.n k/ sgn./ 1:::k 1:::n k D 1;:::;k 1;:::;k X sgn./.1:::k 1:::n k /2 D 1;:::;k X sgn./.1:::k 1:::n k /2 D 1;:::;k kŠsgn./: D In conclusion, we have that

1:::k 1:::n k 1:::n k kŠı : 1:::k 1:::n k 1:::n k D With this, ﬁnally we obtain

1 kŠ 1:::k 1:::n k 1:::n k .?ˇ/ ::: .?ˇ/ ::: ı kŠ.?ˇ/ : 1 n k 1:::k 1:::n k 1 n k 1:::n k 1:::n k .n k/Š D .n k/Š D Now we put it all together: 1:::k kŠ.?ˇ/1:::n k ˇ 1:::k 1:::n k ; D i.e. 1 1:::k .?ˇ/1:::n k ˇ 1:::k 1:::n k : D kŠ And we’re done! Right? We’re done, right? ... guys? What’s wrong? ... What do you mean you want it for a general basis? Okay let u1; : : : ; un be some basis, and let A be the change-of-basis matrix from the e to the u basis, such that ui Ai ej : D j 1Here, the Levi-Civita symbol with lowered indices is not lowered with the metric, i.e. we consider (but only for the Levi-Civita symbol!) 1::: k ::: : D 1 k This means that we are not allowed to raise or lower the indices of with the metric. It is just a convenient symbol for adding things that does not represent the components of a tensor!

5 In this new basis, the volume form is not u1 un, but these two are non-zero top-forms so they’re only ^ ^ scalar multiples of one another. Explicitly,

u1 un A1 :::An e1 en A1 :::An 1:::n e1 en det.A/vol: ^ ^ D 1 n ^ ^ D 1 n ^ ^ D

What is det.A/? Fortunately we can calculate it easily: Let gu be the matrix representation of g on the u-basis, namely Œgu g.u ; u /: D Then T Œgu g.u ; u / A A g.e ; e / A Œge A ŒA ge A ; D D D D where ge is the matrix representation of g with respect to the orthonormal basis, i.e. ge is diagonal with 1 ˙ on the diagonal. Taking the determinant we get

2 s 2 det.gu/ det.A/ det.ge/ . 1/ det.A/ ; D D with s being the number of negative eigenvalues of ge (also known as the signature of the metric). Thus, p det.A/ det.gu/ ; D ˙ j j where the sign depends on the orientation of the u basis. Thus,

1 n p u u det.gu/ vol: ^ ^ D ˙ j j With this we can ﬁnd the expression for ? with respect to any choice of basis. In the u basis, we have 1 1:::k 1:::n k 1 n ˛ ?ˇ ˛1:::k .?ˇ/1:::n k u u ^ D kŠ.n k/Š ^ ^ p det g u 1:::k 1:::n k ˙ j j˛1:::k .?ˇ/1:::n k vol: D kŠ.n k/Š p And thus we can repeat the same process as we did above, just that we need to carry the det.gu/ on the ˙ j j left-hand side for the whole ride. In the end, we get 1 1 .?ˇ/ ˇ1:::k : 1:::n k p 1:::k 1:::n k D ˙kŠ det.gu/ j j Alright before you say anything, yes, I know that this is not the same equation that you’ll see basically everywhere else; the determinant of the metric should in the numerator, you say? Yes, but actually no. See, here we worked with the exterior algebra of a vector space V , not its dual. In practice, with diﬀerential forms, we’re working with the exterior algebra of diﬀerential forms, which are dual to the tangent spaces. That changes the formula a little bit since the matrix of the metric on the dual is the inverse of the matrix in the tangent space.

4 On diﬀerential forms

Before jumping head-ﬁrst to diﬀerential forms, let’s see what happens when we try to apply all this on the dual. If we have a metric g on V , it induces isomorphisms g V V and g] .g / 1 V V , given [ W ! D [ W ! as g .v/.u/ g.v; u/ [ D for all u; v V , and similarly, for all ˛ V , we have that g].˛/ V is such that 2 2 2 g.g].˛/; u/ ˛.u/ D

6 for all u V . If we take a basis (INDEX SWITCH ALERT) u1; : : : ; uk V , and write v v u, then 2 1 k 2 D what are the components of g[.v/? If u ; : : : ; u is the dual basis of V , then we can write

g .v/ .g .v//u ; [ D [ where g .v/ g .v/.u/ g.v; u/ v g.u; u/ v g: [ D [ D D WD Here we write g g.u; u/ as the components of the metric in this basis. We call this lowering the D index of v, and we simply abuse notation by writing g[.v/ vu , with v v g. D WD Similarly, if ˛ ˛u V , then we have that D 2 ] ] ] ˛ ˛.e/ g.g .˛/; e/ .g .˛// g.u; u/ .g .˛// g: D D D D Thus we can invert this matrix equation and write

] 1 .g .˛// .g / ˛; WD 1 2 where .g / are the components of the inverse of the matrix of g. ] So far so good. Now we can simply pull back the metric g from V to V using g , and deﬁne (using the same symbol), for any ˛; ˇ V , 2 g.˛; ˇ/ g.g].˛/; g].ˇ//: D What are the components of the dual metric with respect to the u1; : : : ; un basis? Well we compute

] ] 1 ˛ˇ 1 g g.e ; e / g.g .e /; g .e // .g / .g / g..e / e˛; .e / e/ D D D ˇ 1 ˛ 1 .g / .g / g.e˛; e/ D 1 ˛ 1 .g / .g / g˛ D .g 1/: D

Thus the components of the metric on V are the components of the inverse of the metric on V . We drop the 1 clunky from now on since there is no ambiguity: g always means the components of the inverse of the matrix with entries g. Now we’re done! Let M be a smooth manifold with a Lorentzian metric g. We will apply all this, pointwise, to the cotangent spaces of M . By deﬁnition, g is a smooth tensor ﬁeld on M which is point- dual wise a metric gx TxM TxM R. This metric induces a metric g on TxM via the isomorphism ] W ! .gx/ T M TxM as above. Now we deﬁne the Hodge-dual pointwise but on the cotangent space, so W x ! the metric we use is the inverse of the metric on TM . That is, in the notation of section 2 we let V T M , D x so that the Hodge star is ? k.M / n k.M /, but in this case the components are W x ! x p 1 1 det.gu/ .?ˇ/ ˇ1:::k ˇ1:::k : 1:::n k p 1:::k 1:::n k j j 1:::k 1:::n k D ˙kŠ det.gdual/ D ˙ kŠ j u j If you look at this expression, it is obviously smooth since the components of g and ˇ are and det.gu/ is non-zero. Thus we can happily extend ? to be a global operator

? k.M / n k.M /: W ! 2 1 yes... I know that we don’t carry around the . Just gimme a minute okay?

7 5 A neat example: de Rham vs. curl, grad, div

3 Now let’s make it explicit. We consider M R , with its natural euclidean metric g, and coordinates x; y; z. D The volume form is simply vol dx dy dz: D ^ ^ Now let’s see what ? does to 0, 1, 2, and 3-forms. First, recall that if e1; : : : ; en is an orthonormal basis, then

?.e1 ek / . 1/s1:::k 1:::n k .e1 en k / (no Einstein sum); ^ ^ D ^ ^

1 k where s is the number of elements in e ; : : : ; e with negative length, and 1; : : : ; n k is comple- f g f g mentary to 1; : : : ; k in 1; : : : ; n . Now a 0-form is just a smooth function, say f , and we simply have f g f g ?f f vol f .x; y; z/dx dy dz: D D ^ ^ Now for one-forms, the above result tells us that

?.dx/ dy dz D ^ ?.dy/ dz dx D ^ ?.dz/ dx dy; D ^ so that ?.!xdx !ydy !zdz/ !xdy dz !ydz dx !zdx dy: C C D ^ C ^ C ^ Similarly, for 2-forms we have

?.!yzdy dz !zxdz dx !xydx dy/ !yzdx !zxdy !xydz: ^ C ^ C ^ D C C For 3-forms, ?.f vol/ f: D Now let’s talk about grad, curl, and div. Just as a reminder and for the sake of completeness, let’s write them down. Let U M be an open set. Then we deﬁne grad C .U / X.U / as Â W 1 ! @f @ @f @ @f @ grad.f / g].df /: D @x @x C @y @y C @z @z D We deﬁne curl X.U / X.U /, deﬁned as W ! Â @ @ @ Ã Â@f z @f y Ã @ Â@f x @f z Ã @ Â@f y @f y Ã @ curl f x f y f z ; @x C @y C @z D @y @z @x C @z @x @y C @x @x @z and ﬁnally, div X.U / C .U / given as W ! 1 Â @ @ @ Ã @f x @f y @f z div f x f y f z : @x C @y C @z D @x C @y C @z It is an elementary exercise to prove that curl grad 0 and div curl 0, so that we have a complex ı D ı D grad curl div 0 C .U / X.U / X.U / C .U / 0; ! 1 ! ! ! 1 ! called the gcd complex. Now let’s compare this to the de Rham diﬀerential, explicitly to be more clear: for a 0-form, @f @f @f df dx dy dz: D @x C @y C @z

8 For a 1-form, Â Ã Â Ã Â Ã @!z @!y @!x @!z @!y @!x d.!xdx !ydy !zdz/ dy dz dz dx dx dy: C C D @y @z ^ C @z @x ^ C @x @y ^ And for a 2-form, Â Ã @!yz @!zx @!xy d.!yzdy dz !zxdz dx !xydx dy/ dx dy dz: ^ C ^ C ^ D @x C @y C @z ^ ^ This tells us that there is an isomorphism between the gcd complex and the de Rham complex, given as 1 0 C .U / C .U / being simply 0 id. We also have 1 X.U / .U / as W 1 ! 1 D W ! Â Ã x @ y @ z @ x y z 1 f f f f dx f dy f dz; @x C @y C @z D C C

2 i.e. 1 g . We also have 2 X.U / .U / given as D [ W ! Â Ã x @ y @ z @ x y z 2 f f f f dy dz f dz dx f dx dy; @x C @y C @z D ^ C ^ C ^

3 which we can identify as 2 ? g . Finally, we have 3 C .U / .U / given as D ı [ W 1 !

3.f / f dx dy dz ?.f /: D ^ ^ D

By construction, we have 0 idC .U /, 1 g[, 2 ? g[ and 3 ?, all of which are isomorphisms. D 1 D D ı D Now we need to check that the diagram

grad curl div 0 C 1.U / X.U / X.U / C 1.U / 0

0 1 2 3 0 0.U / d 1.U / d 2.U / d 3.U / d 0 commutes... but this is a straightforward, albeit a bit boring, computation. Therefore we have that is an isomorphism of complexes, which induces an isomorphism in cohomology:

ker.grad/ H 0.U / Š ker.curl/=im.grad/ H 1.U / Š ker.div/=im.curl/ H 2.U / Š C .U /=im.div/ H 3.U /: 1 Š So suppose you have a vector ﬁeld E X.U / satisfying curl.E/ 0. When can you guarantee that 2 D E grad.'/ for some scalar function ' C .U /? The previous result tells us that when H 1.U / 0, i.e. D 2 1 D when U is simply connected, then every irrotational ﬁeld is a gradient. Similarly, if you have a ﬁeld B such that div.B/ 0, then if H 2.U / 0, we can guarantee that B curl.A/ for some ﬁeld A X.U /. D D D 2 6 Another neat example: wedge and cross product

3 Have you noticed that the cross product in R behaves very similarly to the wedge product? With the antisymmetry and all. There is, of course, a huge diﬀerence between both: the cross product returns another 3 3 3 3 vector in R , i.e. R R R , whereas the wedge product returns an element of the exterior product 3 W 3 3 ! 3 3 of R with itself, R R R R . How can we bridge both? ^ W ! ^

9 3 3 3 3 On R we have the canonical euclidean metric g, and thus we have the Hodge star ? R R R . 3 3 3 W ^ ! With this we can construct an antisymmetric map ? R R R . ı ^ W 3 ! Let e1; e2; e3 be the canonical orthonormal basis of R . We then see that

?.e1 e2/ e3 ^ D ?.e2 e3/ e1 ^ D ?.e3 e1/ e2: ^ D However, this is the same as

e1 e2 e3 D e2 e3 e1 D e3 e1 e2: D We then happily conclude that u v ?.u v/ D ^ 3 for all u; v R . 2 7 The takeaway

k n k The Hodge star operator makes an explicit isomorphism between the exterior powers ƒ V and ƒ V of a vector space with the aid of a metric. With it we can bridge the similarities between exterior products of complementary degrees. It also makes explicit the relationship between the cross product and vector calculus 3 in R and the calculus and algebra of diﬀerential forms on it. As a quick corollary, we saw that the existence of potentials for certain functions depends on the topology of the underlying space (which in most cases in physics is trivial). With the Hodge star we will be able to neatly write Maxwell’s equations, and more importantly, generalize them for a large class of physical ﬁelds: gauge ﬁelds.

References

Báez, J. C. and Muniain, J. P. (1994). Gauge Fields, Knots, and Gravity, World Scientiﬁc. Section I.5. Honestly I don’t even know why you’d read this post, go and read Báez instead.

Quintero Vélez, A. (2018). Notas de Fundamentos Matemáticos de las Teorías de Campos Gauge. Fecko, M. (2006). Diﬀerential Geometry and Lie Groups for Physicists. Cambridge University Press. This is a great book with great humor. Chapter 5 is a good introduction to diﬀerential forms. It is a problem-driven book.

Conrad, B. Notes for Math 396: TensorAlgebras, TensorPairings, and Duality.http://virtualmath1. stanford.edu/~conrad/diffgeomPage/handouts/tensor.pdf

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