THE SECOND REAL JOHNSON-WILSON THEORY AND NON-IMMERSIONS OF RP n, PART II NITU KITCHLOO AND W. STEPHEN WILSON Abstract.This paper is a continuation of the study begun in [KW ]. We analyze ER(2)16*+8(RP2n) and compute ER(2)*(RP16K+1) and use these to prove more non-immersion theorems for RPn including many in fairly low dimensions. In particular, we get 12 new non-immersion results for RPn w* *here n 192, the range included in the tables [Dav]. These complement the 10 already found in [KW ]. August 18, 2007 1.Introduction This paper is a continuation of [KW ], which we refer to as Part I. We make * *free use of the notation and results of Part I. The main theorem of [Dav84 ] states that for n = m + ff(m) - 1 k = 2m - ff(m) there does not exist an axial map: K -2k-2 2n 2K -2n-2 RP 2 x RP -! RP and so, by [Jam63 ], RP 2n* R2k for these n and k. This is proven using the equivalent of E(2)*(-) by showing that the u2K-1-n =* * 0 on the right would have to go to a non-zero element on the left. That prevents * *the existence of the axial map. In Part I we constructed a purely algebraic surject* *ion K -2k-4 16* 2K -2k-2 ER(2)16*(RP 2 ) -! E(2) (RP ) that allowed us to show the axial map K -2k-4 2n 2K -2n-2 RP 2 x RP -! RP did not exist if we added the restrictions to k and n that n = 7 or 0 mod 8 and -k - 2 = 1, 2, 5 or 6 mod 8. This improved some non-immersions results by 2. In this paper we are able to include the n = 3 and 4 mod 8 cases by analyzing ER(2)16*+8(RP 2n) and constructing a similar algebraic map K -2k-4 16*+8 2K -2k-2 ER(2)16*+8(RP 2 ) -! E(2) (RP ) with the previous restrictions on k. In order to describe ER(2)16*+8(RP 2n) properly we need to define and study an element y 2 ER(2)8(RP 1). The simple version of our answer, similar to our understanding of ER(2)16*(RP 2n) in Theorem 1.6 of Part I, is: 1 2 NITU KITCHLOO AND W. STEPHEN WILSON Theorem 1.1. A 2-adic basis for ER(2)16*+8(RP 2n) consists of the elements ff2ffkuj, with 0 k and 0 < j < n, yuj, with 0 j n - 4, and, when n = 3 or 4 modulo 8, no other elements, n = 2 or 5 modulo 8, ffkyun-3, n = 1 or 6 modulo 8, ffkyun-3, and yun-2, n = 7 or 0 modulo 8, yun-3, yun-2, and yun-1, and no others. In addition to this we need some information about ER(2)*(-) of products and then we are able to prove: Theorem 1.2. When the pair (m, ff(m)) is, modulo 8, (0, 3),(5, 6), (4, 7) or (1* *, 2), then RP 2(m+ff(m))* R2(2m-ff(m)+1). The most interesting pair to us is (m, ff(m)) = (0, 3). Let m = 8 + 16 + 2i, * *then 2(m + ff(m)) = 54 + 2i+1and 2(2m - ff(m) + 1) = 92 + 2i+2. So, we get i+1 92+2i+2 RP 54+2 * R . The lowest dimensional cases are RP 118* R220 RP 182* R348 However, the importance to us is that it gets on Don Davis's tables, [Dav ]. No* *tice also that for these cases there is now only a knowledge gap of 1 between best k* *nown non-immersions and best known immersions. Next we move on to compute ER(2)*(RP 16K+1), analyze ER(2)8*(RP 16K+1), and construct an algebraic map ER(2)8*(RP 16K+1) -! E(2)8*(RP 16K+2) that allows us to do similar things for non-immersions when, in our axial maps, -k-1 = 1 mod 8. The theory E(2)*(-) cannot make use of the odd spaces because the top cell is not connected algebraically, but for ER(2)*(-) the connection is strong for 16K + 1 and 16K + 9. We have not done the computation for 16K + 9 because, although there are surely more non-immersions there, they are not of l* *ow enough dimension to inspire us to do the work, whereas the 16K + 1 case gives l* *ots of nice new low dimensional results. Theorem 1.3. A 2-adic basis for ER(2)16*(RP 16K+1) is given by the elements ffkuj, 0 k, 0 < j 8K + 1. A 2-adic basis for ER(2)16*+8(RP 16K+1) is given by the elements ff2ffkuj 0 k 0 < j < 8K ff2ffku8K = ffk+1yu8K-3 wffku yuj 0 j < 8K. Using this we get: Theorem 1.4. For the mod 8 pairs (m, ff(m)) = (6, 6), (1, 4), (2, 6), (5, 4) we* * have: RP 2(m+ff(m)-1)* R2(2m-ff(m))+1. NON-IMMERSIONS OF RPn, PART II 3 Let's look at the numbers. First, the pair (1, 4). The lowest possible non- immersions we get from this are i+1 93+2i+2 RP 56+2 * R which also implies another new result: i+1 93+2i+2 RP 57+2 * R . The lowest dimensional examples are: RP 120* R221 RP 121* R221 RP 184* R349 RP 185* R349 The next pair to look at is (5, 4). From this we get i+1+2j+1 13+2i+2+2j+2 RP 16+2 * R . When i = 3 this is: j+1 j+2 RP 32+2 * R45+2 . The lowest dimensional examples are: RP 96* R173 RP 160* R301 When i = 4 this is: j+1 j+2 RP 48+2 * R77+2 . which also implies that j+1 77+2j+2 RP 49+2 * R . The lowest dimensional examples are: RP 112* R205 RP 113* R205 RP 176* R333 RP 177* R333 In the tables, [Dav ], the best known results for non-immersions for RP nfor n 192 are listed. 95 of these are solved completely because it is known that * *RP n immerses in the next higher dimension. Of the remaining 97 cases we improve on 12 in this paper, n = 96, 112, 113, 118, 120, 121, 160, 176, 177, 182, 184, and* * 185, making for a total of 22 when combined with 10 from Part I, n = 48, 62, 80, 94, 110, 126, 144, 158, 174, and 190. The tables also list what is known for n = d + 2i for 0 d < 64. Of these 64 cases, 24 are known completely. Of the remaining 40 we improve on 10, 6 from th* *is paper, d =32, 48, 49, 54, 56, and 57, and 4 from Part I, d =16, 30, 46, and 62. We are fairly confident that ER(2)*(-) will not give any more results in these low dimensions. Before attacking the present cases in this paper, computer com- putations were made on all of the cases we believed we could approach below 192 and we have now proven all of the results that seemed to be there. Thanks to the Fields Institute for Mathematics for supporting the second auth* *or for a week of intense computation (by hand) for this paper. Also, thanks to Mar* *tin Bendersky, Don Davis and Jesus Gonz'alez for their support and inspiration. 2. Injections ER(2)*(X) has already been computed for X = RP 1, RP 2n, and RP 1^RP 1. However, we need a more detailed analysis. Theorem 2.1. The map ER(2)16*+8(RP 1) ! E(2)16*+8(RP 1) is an injection with cokernel given by v42u{1-3}. 4 NITU KITCHLOO AND W. STEPHEN WILSON Proof.Recall that a 2-adic basis for E(2)*(RP 1) is given by vi2ffkuj 0 i < 8 0 k 1 j. The elements of degree 8 mod 16 are v42ffkuj 0 k 1 j. From Theorem 8.1 of Part 1 we can read off the elements of ER(2)16*+8(RP 1). From the x1-torsion we have ff2ffkuj 0 k 1 j. From the x3-torsion we have wffku, k 0, and wuj 1 < j. Mapping these elements to E(2)*(RP 1) we have ff2ffkuj -! 2v42ffkuj = v42ffk+1uj+1 plus higher filtration terms, wffku -! v42ffk+1u and wuj -! v42ffuj = v42uj+2, j > 1 modulo higher terms. From this we can see the injection and that the only terms missed are v42u{1-3}. Theorem 2.2. The map ER(2)16*+8(RP 1 ^ RP 1) ! E(2)16*+8(RP 1 ^ RP 1) is an injection with cokernel given by v42u{1-3}1u{1-3}2. Proof.We have both E(2)*(RP 1^RP 1) and ER(2)*(RP 1^RP 1) written down in Theorem 17.1 of Part I. E(2)*(RP 1 ^ RP 1) is vs2ffkui1u2 0 s < 8 0 k 0 < i vs2ui1uj2 0 s < 8 0 < i 1 < j. The elements in degree 16 * +8 are those with s = 4. We can also write down ER(2)*(RP 1 ^ RP 1) and the map to E(2)*(RP 1 ^ RP 1) in degrees 16 * +8: From the x1-torsion we get ff2ffkui1u2 -! 2v42ffkui1u2 = v42ffk+1ui+11u2 ff2ui1uj2-! 2v42ui1uj2= v42ui+11uj+22 0 < i 1 < j, all modulo higher filtrations. From the x3-torsion we get: wffku1u2 -! v42ffk+1u1u2 0 k wui1u{1,2,3}2-! v42ui+21u{1,2,3}2 1 < i wu1uj2-! v42u1uj+22 1 < j. From this we see the injection and that the only elements missed are those stat* *ed. Note that there are no elements in degrees 16 * +8 divisible by x. Corollaries 8.3 and 17.3 of Part I give us an isomorphism for these spaces in degrees 16* so we get: NON-IMMERSIONS OF RPn, PART II 5 Corollary 2.3. The map ER(2)8*(X) ! E(2)8*(X) is an injection for X = RP 1 and RP 1 ^ RP 1. Not much more work is required to prove: Proposition 2.4. The map ER(2)16*+i(X) ! E(2)16*+i(X) is an injection for X = RP 1 and RP 1 ^ RP 1 when i = 0, 1, 2, 3, 4, 5 and 8. From this we know that there are no elements divisible by x in any of these degrees. 3.y, a new element We need to introduce a new element that we have good control over. We know we have an isomorphism of ER(2)16*(RP 1) and E(2)16*(RP 1) and that we have the same relation, 0 = 2u +F ffu2 +F u4, in both. We can use this to solve for * *u4 as: u4 = -F (2u) -F (ffu2) = 2ug + ffu2h where g and h are invertible power series. As it stands, g and h are not unique* *ly determined, but if we insist that none of the terms of h be divisible by 2 (we * *can move such terms to g) then we can make our choice of g and h unique. Recall that in E(2)*(-) we have set v82= 1. We now multiply this relation, wh* *en viewed only as being in E(2)*(RP 1), by v42, which is a unit in E(2)*(-), to ge* *t a relation v42u4 = v42(2ug + ffu2h) = (2v42)ug + (v42ff)u2h. The image of the element ff2 from ER(2)* in E(2)* is 2v42and the image of w is v42ff so, in this relation, all of the terms on the right hand side are in t* *he image from ER(2)*(RP 1) and we can use them to define a new element y = ff2ug + wu2h that reduces to v42u4 2 E(2)8(RP 1). (The lack of uniqueness of g and h would n* *ot affect anything here. It does not matter whether we convert 2v42to ff2 or v42ff* * to w if we have a v422ff that could be factored either way because ff2ff = 2w 2 ER(2* *)*, [KW07 ].) Although we struggled with this element a great deal in our original computat* *ions and then managed to eliminate it for our work in Part I, it was only with the w* *ork of Bruner, Davis and Mahowald in [BDM02 , DM ] that we realized its importance for our work with non-immersion theorems. The element y has many interesting properties. We collect a few here. Theorem 3.1. There is an element y 2 ER(2)8(RP 1) that maps to v42u4 2 E(2)8(RP 1). We have relations: y2 = u8 ffy = wu4 wy = ffu4 2y = ff2u4 ff2y = 2u4 ff3y = ff1u4 ff1y = ff3u4 xy = xwu2h x3y = 0. Proof.We have already constructed y with the property that it reduces to v42u4. The first five relations take place in degrees 8* where we have an injection of ER(2)8*(RP 1) ! E(2)8*(RP 1) so we can prove the relations by substituting v42u4 for y, 2v42for ff2 and v42ff for w. They all follow quickly then. The n* *ext relation is in degree 4 modulo 16 and we have an injection here too as well from Proposition 2.4 so it also follows by replacing ffi with 2v2i2. Only the next r* *elation 6 NITU KITCHLOO AND W. STEPHEN WILSON requires anything else. It is in degree -4 modulo 16 and we do not have an inje* *ction in this degree. We have to resort to the definition (which we could also have u* *sed for the other relations). ff1y = ff1(ff2ug + wu2h) = (ff1ff2)ug + (ff1w)u2h = (2ff3)ug + (ffff3)u2h = ff3(2ug + ffu2h) = ff3u4. This uses the relations in the coefficient ring: 2ff3 = ff1ff2 and ffff3 = ff1w* *, from [KW07 ]. y = ff2ug + wu2h, so, since xff2 = 0, we get the next relation. Since x3w = 0 the last one follow* *s. We know E(2)*(-) and ER(2)*(-) for RP 1 and RP 1^ RP 1. From Theorem 3.4 of Part I we know that we have a K"unneth theorem for RP 1 ^ RP 1. The standard map RP 1 x RP 1 ! RP 1 induces a coproduct that can be computed from the formal group law, i.e. u ! u1+F u2. However, things are much nicer than that: Theorem 3.2. The coproduct of u, up to a unit, is u1 - u2. The coproduct of y, up to a unit, is y1 - 2ff2u31u2 + 3ff2u21u22- 2ff2u1u32+ y2. Proof.Because we have injections in degrees 8* it is enough to prove this for t* *he image in E(2)*(-). The first statement is well-known and comes from the fact th* *at 0 = [2](X) = X +F X. This implies that X +F Y is divisible by X - Y (because plugging in Y = X gives zero) and so X +F Y = (X - Y )g where g is a power series in X and Y that is invertible, i.e. a unit. Since the coproduct is given* * by the formal group law we have u ! u1 +F u2 = (u1 - u2) up to a unit. To compute the coproduct of y up to a unit we can just compute for u4. Up to a unit this is u41- 4u31u2+ 6u21u22- 4u1u32+ u42. Multiply this by v42and replace* * v42u4i with yi and 2v42with ff2. 4.Rewriting ER(2)*(RP 1) . We would like to rewrite our answer for ER(2)16*+8(RP 1) using our new ele- ment y. Recall, from Theorem 8.1 of Part I, our description of ER(2)*(RP 1). The x1-torsion generators are given by: ffiffkuj 0 i < 4 0 k 1 j where ff0 = 2. The x3-torsion generators are given by: wfflffku, ffl + k > 0, wuj 1 < j, and uj 3 < j. The only x7-torsion generators are u{1-3}. From y = ff2ug + wu2h it is easy to see that we can replace the x3-torsion generators, wuj, 1 < j, using yuj-2. We would also like to replace some of the x1-torsion generators, ff2ffkuj NON-IMMERSIONS OF RPn, PART II 7 with ffk+1yuj-3. This last element is not x1-torsion though. When we are in ER(2)*(RP 2n) and j is big enough, this can be x1 torsion. y = ff2ug + wu2h ff2ug = y - wu2h ff2u = yg-1 - wu2hg-1 ff2u3 = yu2g-1 - wu4hg-1 We know that wu4 = ffy so this is: ff2u3 = yu2g-1 - ffyhg-1. The lead term (i.e. the term with lowest filtration) here is ffy and the whole * *right hand side must be x1-torsion even if the lead term isn't. When the higher filtr* *ation terms are all zero we can replace ff2ffkuj with ffk+1yuj-3. This is enough to give us what we want. 5. ER(2)16*+8(RP 2n) We have, for ER(2)16*+8(-) a theorem similar to Theorem 13.4 of Part I for ER(2)16*(-). Theorem 5.1. For all n > 3 there is a short exact sequence: (5.2) 0 - ER(2)16*+8(RP 2n-2) - ER(2)16*+8(RP 2n) - ER(2)16*+8(RP 2n=RP 2n-2) - 0. We have elements ff2ffkuj 2 ER(2)16*+8(RP 2n), 0 k, 0 < j < n. We also have elements yuj for 0 j n - 4. Depending on n modulo 8 there are other elements in ER(2)16*+8(RP 2n). For n = 8K + 4 and 8K + 3 there are no other elements. For n = 8K +2 there is an x5-torsion element, z16K-38, that reduces to v2u8K+2 in the Bockstein spectral sequence such that x2ffkz16K-38 = ffkyu8K-1. For n = 8K +1 there is an x5-torsion element, z16K-22, that reduces to v2u8K+1 in the Bockstein spectral sequence such that x2ffkz16K-22 = ffkyu8K-2 and an x7-torsion element, z16K-4 that reduces to v62u8K+1 in the Bockstein spe* *ctral sequence such that x2uz16K-22 = x4z16K-4 = yu8K-1. For n = 8K there are x7-torsion elements, z16K-20, and z16K-18 that reduce to v62u8K-1 and v32u8K respectively in the Bockstein spectral sequence such that x4z16K-20 = yu8K-3 x4uz16K-20 = yu8K-2 and x4u2z16K-20 = x6z16K-18 = yu8K-1. For n = 8K + 7 there are x7-torsion elements, z16K-36, and z16K-34 that reduce to v62u8K+6 and v32u8K+7 respectively in the Bockstein spectral sequence such t* *hat x4z16K-36 = yu8K+4 8 NITU KITCHLOO AND W. STEPHEN WILSON x4uz16K-36 = yu8K+5 and x4u2z16K-36 = x6z16K-34 = yu8K+6. For n = 8K + 6 there is an x5-torsion element, z16K-8, that reduces to v2u8K+6 in the Bockstein spectral sequence such that x2ffkz16K-8 = ffkyu8K+3 and an x7-torsion element, z16K-36 that reduces to v62u8K+6 in the Bockstein sp* *ec- tral sequence such that x2uz16K-8 = x4z16K-36 = yu8K+4. For n = 8K +5 there is an x5-torsion element, z16K+10, that reduces to v2u8K+5 in the Bockstein spectral sequence such that x2ffkz16K+10 = ffkyu8K+2. This gives us our Theorem 1.1. Proof.We have computed the Bockstein spectral sequence for all of the spaces RP 2n-2, RP 2n, and RP 2n=RP 2n-2. From this we can just read off the elements in degree 16 * +8. In every case the x1-torsion elements ff2ffkuj for j < n - 1 correspond using the map induced by RP 2n-2! RP 2n. Likewise for the elements wffku, and yuj, 0 j n - 5 so we will ignore these elements. In the proof we* * are constantly using the fact that we already know all of the groups. We also make * *use of the fact that the map ER(2)*(RP 2n=RP 2n-2) ! ER(2)*(RP 2n) was computed explicitly in Equation 13.1 of Part I. First note that ff0ffkun-1 = 2ffkun-1 = ffk+1un. For n = 4 mod 8, there is nothing else in ER(2)16*+8(RP 2n-2). All that is left of 5.2 is ffkz2n 2 ER(2)16*+8(RP 2n=RP 2n-2) and ff2ffkun-1 and yun-4 in ER(2)16*+8(RP 2n). Since there are no elements of higher filtration we can use Section 4 to replace ff2ffkun-1 with ffk+1yun-4. Note that ffkyun-4 is represen* *ted by v42ffkun in the Bockstein spectral sequence. The long exact sequence forces ffkz2n ! ffkyun-4 but so does our direct computation using Equation 13.1 of Part I. For n = 3 mod 8, ER(2)16*+8(RP 2n=RP 2n-2) = 0. We must have ffkyun-4 ! x2ffkun-1v2. [Technically, we need to worry that perhaps yun-4 goes to x2ff3kun* *-1v2 for some k. If this is the case then the boundary homomorphism on x2un-1v2 must be non-trivial but we can check that there is nowhere for it to go. Consequently we will ignore this kind of possibility in the rest of this proof.] For n = 2 mod 8 things are a little more complicated. The only elements in ER(2)16*+8(RP 2n=RP 2n-2) are x2wffkz2n-18 and we can compute directly that they go to x2ffk+1unv2. ffkyun-4 must go to x2ffkun-1v2. The only possibility l* *eft is for x2unv2 to go to x4un-1v62. Recall from above that this last element is y* *un-3. For n = 1 mod 8, we compute the map to ER(2)16*+8(RP 2n) directly and we have wffkz2n-18- ! ffk+1yun-4 x2wffkz2n -! x2ffk+1unv2 Keep in mind that this last represents ffk+1yun-3. yun-4 must map to x4v62un-2, x2v2un = yun-3 to x4v62un-1, and x4v62un = yun-2 to x6v32un-1. NON-IMMERSIONS OF RPn, PART II 9 For n = 0 mod 8 we compute x6z2n-18 ! x6unv32= yun-1 and wffkz2n ! ffk+1yun-4. That leaves yun-4 ! x4un-2v62, x4un-1v62= yun-3 ! x4un-1v62, and x4unv22= yun-2 ! x6un-1v32. For n = 7 mod 8 we compute x4z2n-18 ! x4unv62= yun-2 and x6z2n ! x6unv32= yun-1. That leaves x4un-1v22= yun-3 ! x4un-1v62and ffkyun-2 ! x2ffkun-1v2. For n = 6 mod 8 we compute x2ffkz2n-18! x2ffkunv2 = ffkyun-3 and x4z2n ! x4unv62= yun-2. All that is left is ffkyun-4 ! x2ffkun-1v2. The n = 5 mod 8 case is simple again with ffkz2n-18! ffkyun-4 and x2ffkz2n ! x2ffkunv2 = ffkyun-3. 6.Algebraic maps We can now see, from Theorem 5.1 and Part I, that we have purely algebraic maps, no topology used or implied, of ER(2)8*(RP 2n) -! E(2)8*(RP 2n+2) n = 1, 2, 5, 6 mod 8. These maps are neither injective nor surjective. However, they are close enough to surjective for our purposes since the only elements they miss are the v42u{1* *-3}. These low powers of u are never involved with our non-immersion results. 7. Last of the even spaces The goal of this section is to prove Theorem 1.2. We begin again with the main theorem of [Dav84 ], for n = m + ff(m) - 1 k = 2m - ff(m) there does not exist an axial map: K -2k-2 2K -2n-2 RP 2nx RP 2 -! RP and so RP 2n* R2k. This is proven by using the equivalent of E(2)*(-) and showing that the u2K-1-n = 0 on the right would have to go to a non-zero element on the left. That same element would prevent the existence of an axial map, K -2k-2 2K -2n-2 RP 2n+2x RP 2 -! RP and likewise K K RP 2n+2x RP 2 -2k-2-! RP 2 -2n-4. Furthermore, if u2K-1-n went to non-zero then we must also have u2K-1-n-1 = 0 goingKto-a1non-zero element. IfKn + 1 = 3 mod 8 then, from Theorem 1.1, we know yu2 -n-5 = 0 in ER(2)*(RP 2 -2n-4) and, if -k - 2 = {1, 2, 5, 6} mod 8 we have a commuting diagram: (7.1) E(2)*(RP 2n+2) E(2)*E(2)*(RPO2KO-2k-2)oo___ E(2)*(RPO2KO-2n-4) | | | | | | ER(2)*(RP 2n+2) ER(2)*ER(2)*(RP 2K -2k-4)oo_ ER(2)*(RP 2K -2n-4). Our element yu2K-1-n-5 in the lower right hand corner maps to v42u2K-1-n-1 in the upper right hand corner. v42is a unit so the result of Davis shows that 10 NITU KITCHLOO AND W. STEPHEN WILSON thisKelement-maps1non-trivially to the upper left hand side of our diagram and * *so yu2 -n-5 must map non-trivially to the lower left hand side. If we show that the corresponding element in the tensor product maps non- trivially to ER(2)*(-) of the product, then we will have the following proposit* *ion: Proposition 7.2. When n = m + ff(m) - 1 and k = 2m - ff(m), with n + 1 = 3 mod 8 and -k - 2 = {1, 2, 5, 6} mod 8, there is no axial map K -2k-4 2K -2n-4 RP 2n+2x RP 2 -! RP . and so RP 2n+2* R2k+2. To derive the proof of Theorem 1.2 from this we have to untangle some equatio* *ns to get our (m, ff(m)) pairs. We have n + 1 = m + ff(m) = 3 mod 8 and -k - 2 = -2m + ff(m) - 2 = {1, 2, 5, 6} mod 8. The equation for k gives 2m - ff(m) = {5, 4, 1, 0} mod 8. The equation for n gives m + ff(m) = 3 mod 8. Adding, we have 3m = {0, 7, 4, 3} mod 8. Multiply by 3 to get m = {0, 5, 4, 1} mod 8. Substituting into ff(m) = -m + 3 mod 8 we get ff(m) = {3, 6, 7, 2} and this gives us (m, ff(m)) pairs (0, 3),(5, 6), (4, 7) and (1, 2) mod 8. To complete our proof of Theorem 1.2 we have to show that the obstruction which we know is non-zero in our tensor product actually maps non-trivially to * *the ER(2)*(-) of the product. We do not need to know much about the actual obstruction. It is enough to know that it must be a linear combination of our 2-adic basis for the tensor product* * in the upper left hand corner of our diagram. The basis here, in degrees 16 * +8, is g* *iven by v42ffkui1u2 with i n + 1 and v42ui1uj2with i n + 1 and 1 < j 2K-1 - k * *- 1. We need to discuss the obstruction just a little in order to be carefulKwith-* *our1 comparison. Recall from Theorem 3.2 that computing the coproduct of v42u2 -n-1 is, up to a unit, (u1 - u2)2K-1-n-1. The algorithm, Remark 15.3 of Part I, never lowers the powers of the u's so the obstruction must be a linear combination of the 2-adic basis elements v42ui1uj2with i n + 1, 1 < j 2K-1 - k - 1, and i + j 2K-1 - n - 1. (The missing v42u{1,2,3}1u{1,2,3}2never figure in here.) We need to show that the elements which hit these from ER(2)*(-) map non- trivially from the tensor product: K -2k-4 (7.3) ER(2)*(RP 2n+2) ER(2)*ER(2)*(RP 2 ) NON-IMMERSIONS OF RPn, PART II 11 K -2k-4 -! ER(2)*(RP 2n+2x RP 2 ). This is easy for most of these elements. Consider the elements ui1uj2y2 with 3 * *< i n+1 and for 3 < j 2K-1 -k -6. These reduce non-trivially (and independently) to v42ui1uj+42in E(2)8*(-). That's not quite enough though. We also need the K-1-k-5 elements ui1u22 y2 for 3 < i n + 1. The following result finishes off what we need. Proposition 7.4. When q = 1, 2, 5 or 6 modulo 8 and m 8K and 8K + 8 < q, the element um1uq-32y2 2 ER(2)*(RP 2m^ RP 2q) is non-zero. Proof.The element yuq-3 is represented in the spectral sequence for ER(2)*(RP 2* *q) by x2v2uq (from Theorem 5.1) so the element um1uq-32y2 is represented by x2um1v* *2uq2. Thus it is enough to show that the element v2um1uq2survives in the spectral seq* *uence to E3. In Theorem 19.2 of Part I we have computed the entire E2 term of the spectral sequence and this term is there with no restrictions on q. There can be no differentials on this element since it is the product of two honest elements* * (this does use the restriction on q). All we need to do now is show this element is not in the image of d2. The differential d2 has degree 35 = -13. Our element um1uq2v2 has degree -16(m+q)-6 so the source that would have to hit it would have to have degree -16(m + q) - * *41, in particular, it must be odd degree. From Theorem 19.2 of Part I, the odd degr* *ee elements in the E2 term of our Bockstein spectral sequence are: v2s2ffkz-16q-17 v2s2ui1z-16q-17 0 < i < m and v2s+12ffkum-11z-16q-17 The only elements with degree equal to -9 modulo 16 are: v42ffkz-16q-17 and v42ui1z-16q-17 0 < i < m. In the proof of Proposition 19.3 of Part I we showed that d2 must be trivial on z-16q-17when m 8K. d2 also commutes with multiplication by ff and u1. We also know that since d2(v42) = 0, d2 commutes with multiplication by v42. Consequent* *ly, d2 is trivial on all of the elements listed above when m 8K. This completes the proof of Theorem 1.2. 8. ER(2)*(RP 16K+1) We begin our computation by setting up the Bockstein spectral sequence. E1 is just E(2)*(RP 16K+1), which is nothing more than E(2)*(RP 16K) E(2)*(S16K+1). So, we have a 2-adic basis, (it isn't really necessary to use this notation for* * the torsion free part and so it isn't necessary to go to the 2-adic completion of E* *R(2), it is just convenient notation now), letting 2 = ff0, E1 vi2ffkuj 0 i < 8 0 k 0 < j 8K. vi2ffq0ffk'16K+1 0 i < 8 0 q 0 k. 12 NITU KITCHLOO AND W. STEPHEN WILSON All of the first part is even degree and all of the second part is odd degree* *. d1 is even degree so it is induced by the maps RP 16K-! RP 16K+1-! S16K+1 where we already know it. Thus we can just read off our d1 from Theorem 13.2 Part I, for the RP 16Kpart and Section 5 Part I for the S16K+1 part. d1(v2s-52ffkuj) = 2v2s2ffkuj = v2s2ffk+1uj+1 j < 8K (modulo higher powers of u). d1(ffq0v2s+12ffk'16K+1) = ffq+10v2s-22ffk'16K+1 E2 is given by: v2s2ffku0 k, v2s2uj1 < j 8K, v2s+12ffku8K0 k v2s2ffk'16K+1. We confront a new problem now. d2 has degree 35 and we have both odd and even degree elements so it could be non-zero. If so, by naturality it must have its source in the RP 16Kpart and its target in the S16K+1 part. Furthermore the source cannot be something in the image from the E2 for ER(2)*(RP 1) because we know that d2 is zero on all of those elements. All we are left with for poss* *ible sources is v2s+12ffku8K 0 k The differential d2 is trivial on v22and ff so it commutes with multiplication * *by these elements. Since v22is a unit, if there is a d2 then it must be non-zero on v2u8* *K which has degree -6 - 16(8K) = 16K - 6. The degree of the target must be this plus 35, or 16K + 29 = 16K - 19. The possible targets have degrees -12s - 32k + 16K + 1. Working modulo 16 we need -3 = -12s + 1 so we see that s = 3 and we have 16K - 19 = -36 - 32k + 16K + 1, which gives -19 = -36 - 32k + 1 modulo 48. This is 16 = -32k which suggests the solution of k = 1 (other alternatives are k = 3q + 1). If there is a d2 we would conjecture that it starts with d2(v2u8K ) = v62ff'16K+1 and this would lead to d2(v2s+12ffku8K ) = v6+2s2ffk+1'16K+1. We now know what to look for. If we can show that the element ff'16K+1 is in the image from S16K+1 and that x2 times it must be zero, then our conjectured d2 is correct. If we look carefully at ER(2)*(S16K+1) we see that there are a number of known 2-torsion free elements, namely, all of wfflffk'16K+1, ff{1,3}ffk'16K+1 and ff2* *'16K+1. If we look at RP 16K-! RP 16K+1-! S16K+1 we know that ER(2)*(RP 16K) is all torsion so our torsion free elements must all inject into ER(2)*(RP 16K+1). From this we know that the element ff'16K+1 is in ER(2)*(RP 16K+1), but we don't yet know if x2 kills it. For that we need the diagram: NON-IMMERSIONS OF RPn, PART II 13 (8.1) RP 16K______=______//RP 16K | | | | | | | | fflffl| fflffl| RP 16K+1___________//RP 16K+2_________//_S16K+2 | | | | | | | | =| | | | fflffl| fflffl| fflffl| S16K+1 _______//_RP 16K+2=RP 16K_____//S16K+2 Each row and column is a cofibration giving rise to a long exact sequence. Our goal is to show that x2ff'16K+1 in ER(2)*(RP 16K+1) is zero. It is the image of* * the same named element in ER(2)*(S16K+1). From Corollary 9.3 of Part I we know that the element z16K-16 2 ER(2)*(RP 16K+2=RP 16K) maps to x'16K+1 in ER(2)*(S16K+1). Consequently, xffz16K-16 must map to our element of interest, x2ff'16K+1. Rather than go through S16K+1 on our way to RP 16K+1we can now go through RP 16K+2. The element z16K-16 maps to u8K+1 (Equation 13.1 Part I) and so xffz16K-16 maps to xffu8K+1. Since 2x = 0 we can use the relation (Equation 1.3 Part I) on ffu2 and we will get xu8K+3 plus even higher terms, but we know that u8K+3 is zero in ER(2)*(RP 16K+2) (Theorem 1.6 Part I). So, it follows that we have x2ff'16K+1 = 0 in ER(2)*(RP 16K+1) and we can compute our d2 as we conjectured. We get more information out of that computation. It shows us that u8K+1 is represented in the spectral sequence by x'16K+1 because both come from z16K-16. Multiply u8K by ff2 and in E(2)*(RP 16K+2) this is v42ffu8K+1. Consequently, ff2ffku8K is represented by v42ffk+1u8K+1 and in RP 16K+1this is represented by v42ffk+1x'16K+1. From our discussion in Section 4 we can replace this ff2ffku8K* * with ffk+1yu8K-3. We now have our E3: v2s2ffku0 k, v2s2uj1 < j 8K, v2s2'16K+1. d3 is even degree again so the even and odd degree parts don't mix. In S16K+1 d3 takes v22to ffv42but this element is not there, so there is no d3 on the odd* * part. On the even part we already know the d3 differentials: d3(v{6,2}2ffku) = v{0,4}2ffk+1u d3(v{6,2}2uj) = v{0,4}2ffuj = v{0,4}2uj+2 1 < j 8K - 2 We get: E4: v{0,4}2u{1-3}, v{6,2}2u{8K-1,8K}, v2s2'16K+1. 14 NITU KITCHLOO AND W. STEPHEN WILSON Our d4 is odd degree again and so must go from the RP 16Kpart to the S16K+1 part if at all. d4 has degree 21 and must be zero on things in the image from R* *P 1 so a non-zero differential must start out on v{6,2}2u8K-1 and hit one of v2s2'16K+1. The source degrees are -36 - 16(8K - 1) = 16K - 20 and 16K + 4. Adding 21 to see what degree our target would have to be we get 16K + 1 and 16K + 25. Since v42commutes with d4, if we have a d4 it must be d4(v62u8K-1) = '16K+1 d4(v22u8K-1) = v42'16K+1. If x4'16K+1 = 0 then this must be the case. The argument here is the same as before using the diagram 8.1. z8K-16 maps to x'16K+1 in S16K+1 so we want to study x3z8K-16 and we will again get to RP 16K+1 by way of RP 16K+2where x3z16K-16 maps to x3u8K+1 = 0 (because x3u4 = 0). This computes our d4 and we have left for our E5: v{0,4}2u{1-3}, v{6,2}2u8K , v{2,6}2'16K+1. d5 is once again even degree and the dimensions don't work for the odd part so it is zero there and RP 8Kdetermines it is zero on the even part. We have to consider the possibility of a d6 which is of degree 7. Again, it m* *ust go from even to odd by naturality. It would commute with v42so if d6 is non-zero on v22u8K it will be non-zero on the other element. The degree here is -12 + 16* *K. Add 7 to look for the target to get -5 + 16K. Elements that are in odd degrees are in degrees 16K + 1 - 12 and 16K + 1 - 36 so there can be no d6. All we have left is d7 and we know, for starters, that d7(v42u1-3) = u1-3. We can also read off from RP 8Kthat d7(v22u8K ) = v62u8K . The only issue remaining is how d7 works on v{2,6}2'16K+1. The element yu8K-1 is represented by x4v62u8K+1 in the Bockstein spectral se- quence for ER(2)*(RP 16K+2) and by x6v32u8K in the Bockstein spectral sequence for ER(2)*(RP 16K) (Theorem 5.1). It must pass through ER(2)*(RP 16K+1) non-trivially and it must be divisible * *by x4 in here. The only even degree candidates for such an element in the Bockstein spectral sequence for ER(2)*(RP 16K+1) are x{4,6}v62u8K and x5v{2,6}2'16K+1. The degree of yu8K-1 is 16K 24 which is 8 mod 16. Checking x{4,6}v62u8K and x5v{2,6}2'16K+1 we see that, modulo 16, their degrees are 8, -10, 0, and 8 resp* *ectively so the only possibilities are x4v62u8K and x5v62'16K+1. However, modulo 48, the* *se are -20 - 36 + 16K and -36 - 36 + 16K, or, 16K - 8 and 16K - 24 so it must be represented by x5v62'16K+1 so our last undecided differential must be d7(v22'16* *K+1) = v62'16K+1. This concludes our computation of ER(2)*(RP 16K+1) using the Bockstein spec- tral sequence and we collect our results here. Theorem 8.2. The Bockstein spectral sequence for ER(2)*(RP 16K+1) is as fol- lows. NON-IMMERSIONS OF RPn, PART II 15 E1 vi2ffkuj 0 i < 8 0 k 0 < j 8K. vi2ffq0ffk'16K+1 0 i < 8 0 q 0 k. d1(v2s-52ffkuj) = 2v2s2ffkuj = v2s2ffk+1uj+1 j < 8K (modulo higher powers of u). d1(ffq0v2s+12ffk'16K+1) = ffq+10v2s-22ffk'16K+1 E2 v2s2ffku0 k, v2s2uj1 < j 8K, v2s+12ffku8K0 k v2s2ffk'16K+1. d2(v2s+12ffku8K ) = v6+2s2ffk+1'16K+1. E3: v2s2ffku0 k, v2s2uj1 < j 8K, v2s2'16K+1. d3(v{6,2}2ffku) = v{0,4}2ffk+1u d3(v{6,2}2uj) = v{0,4}2ffuj = v{0,4}2uj+2 1 < j 8K - 2 E4: v{0,4}2u{1-3}, v{6,2}2u{8K-1,8K}, v2s2'16K+1. d4(v{6,2}2u8K-1) = v{0,4}2'16K+1. E5 = E6 = E7: v{0,4}2u{1-3}, v{6,2}2u8K , v{2,6}2'16K+1. d7(v42u{1-3}) = u{1-3} d7(v22u8K ) = v62u8K d7(v22'16K+1) = v62'16K+1 We identify all of the elements in degree 8*. This completes the proof of The* *orem 1.3. Theorem 8.3. A 2-adic basis for the elements in ER(2)8*(RP 16K+1) is given by the following. From the x1-torsion elements we have ffkuj 0 < k 1 < j 8K represents elements with the same name. Also ff2ffkuj 0 k 0 < j < 8K is represented by 2v42ffkuj = v42ffk+1uj+1 16 NITU KITCHLOO AND W. STEPHEN WILSON modulo higher powers of u. From the x2-torsion we have ffku8K+1 k > 0 is represented by ffkx'16K+1. and ff2ffku8K = ffk+1yu8K-3 is represented by v42ffk+1x'16K+1. From the x3-torsion we have ffk+1u represents the element with the same name and v42ffk+1u represents wffku. uj 3 < j 8K represents the element with the same name and v42uj 3 < j 8K represents yuj-4. From the x4-torsion we have x'16K+1 represents u8K+1 and v42x'16K+1 represents yu8K-3. From the x7-torsion we have u{1-3} represent elements of the same name. x4v62u8K represents yu8K-2. and x5v62'16K+1 represents yu8K-1. NON-IMMERSIONS OF RPn, PART II 17 Proof.We can find all of the elements in degrees 8* by looking at the Bockstein spectral sequence. If we check the elements that are x1-torsion, i.e. the image* * of d1, the elements in degrees 8* are just 2v{0,4}2ffkuj, and these are, modulo hi* *gher filtrations, v{0,4}2ffk+1uj+1, 0 < j < 8K. These can be written as ffiffkuj wit* *h i = 0 and 2, 0 < j < 8K. Elements in degree 8* coming from the x2 torsion are represented by x times v{0,4}2ffk+1'16K+1. In our computation of d2 in the Bockstein spectral sequence* * we showed that z16K-16 mapped to x'16K+1. We also showed it mapped to u8K+1. This was only x2 torsion if we multiplied by ffk+1. This gives us our ffk+1u8K+1 which is 2ffku8K . From the proof of Theorem 5.1 we know that wffkz16K-16 maps to ffk+1yu8K-3 and is represented by v42ffk+1u8K+1, or, x times v42ffk+1'16K+1. This identifie* *s our remaining 8* degree elements coming from the x2 torsion. The only elements that come from the x3-torsion are again standard elements, wfflffku with ffl + k > 0, uj with 3 < j 8K, and yuj with 0 j 8K - 4. From the x4-torsion we get xv{0,4}2'16K+1 in the expected degrees. We have already identified each of these as u8K+1 and yu8K-3 respectively. Of course our x7-torsion u{1-3}is standard. We have only the x7-torsion elements v62u8K and v62'16K+1 remaining to consid* *er. The only elements in the appropriate degrees are x4 times the first and x5 time* *s the second. We have already identified the last one as yu8K-1. We have not identifi* *ed the necessary element yu8K-2 which we now see must be x4v62u8K . We have a Corollary: Corollary 8.4. There is a purely algebraic map ER(2)8*(RP 16K+1) -! E(2)8*(RP 16K+2) which only misses the elements v42u{1-3}. 9. axial maps and odd spaces Recall that Don Davis uses E(2)*(-) to show that the axial map K -2k-2 2n 2K -2n-2 RP 2 x RP -! RP does not exist when n = 2(m + ff(m) - 1) and k = 2(2m - ff(m)) giving him RP 2(m+ff(m)-1)* R2(2m-ff(m)). From the previous section we know that there is an algebraic map, which, for our purposes, is surjective enough, when -2k - 2 = 2 mod 16: K -2k-3 * 2K -2k-2 ER(2)*(RP 2 ) -! ER(2) (RP ) We will be able to use our standard tricks to show that RP 2(m+ff(m)-1)* R2(2m-ff(m))+1. when -k - 1 = 1 = -2m + ff(m) - 1 mod 8 and m + ff(m) - 1 = n = {3, 4, 7, 0}. 18 NITU KITCHLOO AND W. STEPHEN WILSON Presumably one could compute ER(2)*(RP 16K+9) and prove a similar theorem with -k - 1 = 5. Although the results are probably new they are not of sufficie* *ntly low dimensions as to interest us. Before we proceed let's check out the numbers here. We have 2m - ff(m) = -2 mod 8 m + ff(m) = {4, 5, 0, 1} Adding we get 3m = {2, 3, 6, 7}. Multiply by 3 (always mod 8) m = {6, 1, 2, 5}. Then ff(m) = {6, 4, 6, 4}. This is Theorem 1.4 in the introduction. The rest of the paper is dedicated to the proof. 10.The 16* cases. We now have the diagram in degrees 16*: (10.1) E(2)*(RP 2n) E(2)*E(2)*(RPO2KO-2k-2)oo___ E(2)*(RPO2KO-2n-2) | | | | | | ER(2)*(RP 2n) ER(2)*ER(2)*(RP 2K -2k-3)oo_ ER(2)*(RP 2K -2n-2). We assume throughout that 2K - 2k - 3 is equal to 1 mod 16. In the case of n = 7 or 0 mod 8, we have isomorphisms in degrees 16* on the f* *ar left term and the right term, and, in the relevant degrees, a surjection in the* * right side of the tensor product.K-1 We use the fact that u2 -n on the right is zero. Consequently, with the Don Davis's restrictions on n and k we know our ob- struction is non-zero in the tensor product. All we have to do now is show that* * the elements corresponding to that tensor product also exist in K -2k-3 ER(2)16*(RP 2n^ RP 2 ). As in previous cases, because the algorithm always increases the number of u's * *and because the coproduct of u can be computed as u1 - u2 up to a unit, we just need the elements ui1uj2to be non-zero when i + j is big. (We don't really have to w* *orry about the ffui1u2 terms because they don't have enough u's in them.) Most of the ui1uj2are obviously non-zero and independent because they reduce to E(2)*(-). The only elements this doesn't work for are taken care of by the following theorem. Theorem 10.2. When n 8M < 8M + 8 < 8K, in ER(2)16*(RP 2n^ RP 16K+1). the element un1u8K+12is non-zero. NON-IMMERSIONS OF RPn, PART II 19 If this element is non-zero, since the elements ui1u8K+12are defined and un-i1 times them is non-zero, they too are all non-zero. This result will complete the proof of the theorem in the previous section fo* *r the n = 7, 0 cases. 11.products with an odd space We study the Bockstein spectral sequence for ER(2)*(RP 2n^ RP 16K+1). where 2n < 16K + 1. The E1 term is, as usual, just E(2)*(RP 2n^ RP 16K+1). This has a few more pieces than we are used to. Because E(2)*(RP 16K+1) ' E(2)*(RP 16K) E(2)*(S16K+1). Since E(2)*(S16K+1) is free it doesn't affect the Tor term, only the tensor pro* *duct term. So our E1 is E(2)*(RP 2n) E(2)*(RP 16K) E(2)*(RP 2n) E(2)*(S16K+1) -16(8K)-1E(2)*(RP 2n) Keep in mind that E(2)*(RP 2n) E(2)*(S16K+1) ' 16K+1E(2)*(RP 2n) The -16(8K)-1 looks silly and can be replace with 16K-1 since we are working modulo 48. We need our 2-adic basis for our E1 term. vs2ffkui1u2 0 k 0 < i n s < 8 vs2ui1uj2 0 < i m 1 < j 8K s < 8 and vs2ffkui1'16K+1 0 k 0 < i n s < 8 vs2ffkui1z16K-17 0 k 0 i < n s < 8 We know that x'16K+1 represents u8K+12so xun1'16K+1 represents un1u8K+12. There can be no differential on un1'16K+1 because it is a product of elements. * *All we have to do is show that it is not in the image of d1. Since d1 is even degr* *ee we only have to worry about it on the odd degree elements since un1'16K+1 is odd degree. d1 has degree 18 so if d1 is to hit un1'16K+1 it must start on some ffkui1z16* *K-17 because they are the only elements in the appropriate degree mod 16. Since d1 commutes with ff we would have to have ui1z16K-17 hitting un1'16K+1 and since d1 commutes with multiplication by u1 we would have to have d1 be non-trivial on z16K-17. In the Bockstein spectral sequence for ER(2)*(RP 16M+16xRP 16K+2), 8M +8 < 8K we have, from Theorem 19.2 of Part I, that d1(z16K-33) = 0. From Theorem 1.2 of [GW ], z16K-33 maps to u1z16K-17 in the spectral sequence for RP 16M+16x RP 16K. Since this passes through the spectral sequence for RP 16M+16xRP 16K+1, z16K-33 maps to u1z16K-17 here as well so d1(u1z16K-17) = 0, i.e. u1 multiplied times d1(z16K-17) is zero. All elements killed by multiplication by u1 go to ze* *ro 20 NITU KITCHLOO AND W. STEPHEN WILSON under the map to RP 8Mx RP 16K+1, and so, in here, our d1(z16K-17) = 0 and our result follows by naturality. This concludes our proof for the n = 7, 0 cases. 12. The 16 * +8 cases. Again, we have the diagram when each factor is restricted to degrees 8*: (12.1) E(2)*(RP 2n) E(2)*E(2)*(RPO2KO-2k-2)oo___ E(2)*(RPO2KO-2n-2) | | | | | | ER(2)*(RP 2n) ER(2)*ER(2)*(RP 2K -2k-3)oo_ ER(2)*(RP 2K -2n-2). We continue to assume that 2K - 2k - 3 is equal to 1 mod 16. K-1 We now switch to the cases of n = 3,K4-mod18. We now have yu2 -n-4 = 0 on the lower right which maps to v42u2 -n = 0 in the upper right. [Dav84 ] sh* *ows this goes to non-zero in the upper left hand corner. In the tensor product we have surjections for degrees 8* in both factors for * *high powers of ui. The discussion now is nearly identical to that for the even products we studi* *ed first in this paper. The end result that we need to complete the work is: Theorem 12.2. When m 8M and 8M + 8 < 8K the element um1u8K-32y2 2 ER(2)*(RP 2m^ RP 16K+1) is non-zero. We have already written down the E1 term for this. The element that represents um1u8K-32y2 is v42xum1'16K+1. All we have to do is show that v42um1'16K+1 is no* *t the target of a d1. If there is such a differential for 8M + 8, just like in the la* *st case, d1(z-16n-17) must be non-zero and u1 times the target must be zero. All such target elements go to zero when we map down to m 8M < 8M + 8 and so the differential is trivial there. This concludes the proof of the final cases. References [BDM02]R. R. Bruner, D. M. Davis, and M. Mahowald. Nonimmersions of real projec* *tive spaces implied by tmf. In D. Davis, J. Morava, G. Nishida, W. S. Wilson, and N.* * Yagita, editors, Recent Progress in Homotopy Theory: Proceedings of a conference* * on Recent Progress in Homotopy Theory March 17-27, 2000, Johns Hopkins University,* * Balti- more, MD., volume 293 of Contemporary Mathematics, pages 45-68, Providen* *ce, Rhode Island, 2002. American Mathematical Society. [Dav] D. Davis. Table of immersions and embeddings of real projective* * spaces. http://www.lehigh.edu/~dmd1/immtable. [Dav84]D. M. Davis. A strong nonimmersion theorem for real projective spaces. A* *nnals of Mathematics, 120:517-528, 1984. [DM] D. M. Davis and M. Mahowald. Nonimmersions of real projective spaces imp* *lied by tmf, revisited. Preprint. [GW] J. Gonz'alez and W.S. Wilson. The BP-theory of two-fold products of proj* *ective spaces. In preparation. NON-IMMERSIONS OF RPn, PART II 21 [Jam63]I.M. James. On the immersion problem for real projective spaces. Bulleti* *n of the Amer- ican Mathematical Society, 69:231-238, 1963. [KW] N. Kitchloo and W.S. Wilson. The second real Johnson-Wilson theory and n* *on- immersions of RPn. Manuscript. [KW07] N. Kitchloo and W.S. Wilson. On the Hopf ring for ER(n). Topology and it* *s Applica- tions, 154:1608-1640, 2007. Department of Mathematics, University of California, San Diego (UCSD), La Jol* *la, CA 92093-0112 E-mail address: nitu@math.ucsd.edu Department of Mathematics, Johns Hopkins University, Baltimore, Maryland 21218 E-mail address: wsw@math.jhu.edu