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 / .