Seminar XXIV Final Sessions 1 - Lacan in Ireland

Seminar XXIV Final Sessions 1 - Lacan in Ireland Seminar XXIV Final Sessions 1 - Lacan in Ireland

lacaninireland.com
from lacaninireland.com More from this publisher
10.07.2015 Views

proceeds also to what I would call a flattening out for this to be clear. Theflattening out which on this occasion is spherical is necessary for one to put one’sfinger on the fact, as I might say, that the crossovers in question, the tetrahedriccrossovers, are indeed of the same order, namely, that the tetrahedron which isunderneath, the third tetrahedron, passes underneath, and that the tetrahedronwhich is above, the third tetrahedron passes above. It is indeed because of thatthat we are still here dealing with the Borromean knot.What is annoying nevertheless, is that even in space, even starting from apresupposed spatial, we should also be constrained in this case here to support –since when all is said and done, it is we who support it – to support the flatteningout. Even starting from a spatial presupposition, we are forced to support thisflattening out, very precisely in the form of something which presents itself as asphere (Fig. V-5b). But what does that mean, if not, that even when wemanipulate space, we have never seen anything but surfaces, surfaces no doubtwhich are not banal surfaces because we articulate them as flattened out. Fromthat moment on, it is manifest on the balls that the fundamental plait, the onethat crisscrosses itself 12 times, it is manifest that this fundamental plait formspart of a torus. Exactly this torus that we can materialise by the following,namely, the twelve-fold plait, and that we can also moreover materialise in termsof the following namely, the six-fold plait [Fig. V-3 and Fig. V-4].In truth this function of torus is clearly manifest in the balls that I have just givenyou, because it is no less true that between the two little triangles, if we make – Iwould ask you to consider these balls – if we make a polar thread pass through,we will have exactly in the same way a torus; for it is enough to make one hole atthe level of these two little triangles to constitute at the same time a torus. Thisindeed is why the situation is homogenous, in the case of the Borromean knot, asI have drawn it here, is homogenous between the Borromean knot and thetetrahedron.There is therefore something which ensures that it is no less true for atetrahedron that the function of the torus governs here whatever is nodal in theBorromean knot. It is a fact, and it is a fact that has strictly never been glimpsed,52

namely, that everything that concerns the Borromean knot is only articulated bybeing toric.A torus is characterised quite specifically as being one hole. What is annoying, isthat this hole is difficult to define. The fact is that the knot of the hole with itsflattening out is essential, it is the only principle of their counting – and that thereis only one way, up to the present, in mathematics, of counting the holes: it is bygoing through, namely, by taking a path such that the holes are counted. This iswhat is called the fundamental group. This indeed is why mathematics does notfully master what is at stake.How many holes are there in a Borromean knot? This indeed is what isproblematic since, as you see, flattened out, there are four of them [Fig. V-6].There are four of them, namely, that there are not fewer than in the tetrahedronwhich has four faces in each of the faces of which one can make a hole. Exceptfor the fact that one can make two holes, even three, even four, by making a holein each of these faces and that, in this case, each face being combined with all theothers and even repassing through itself, it is hard to see how to count thesepaths which would be constitutive of what is called the fundamental group. Weare therefore reduced to the constancy of each of these holes which, by this veryfact, vanishes in a quite tangible way, since a hole is no great thing.How then distinguish what makes a hole and what does not make a hole?Perhaps the quatresse can help us to grasp it.What is involved in this quatresse is something which solidarises what is found,that by which it happens that I qualified three circles, namely, that, as you see53

proceeds also to what I would call a flatten<strong>in</strong>g out for this to be clear. Theflatten<strong>in</strong>g out which on this occasion is spherical is necessary for one to put one’sf<strong>in</strong>ger on the fact, as I might say, that the crossovers <strong>in</strong> question, the tetrahedriccrossovers, are <strong>in</strong>deed of the same order, namely, that the tetrahedron which isunderneath, the third tetrahedron, passes underneath, and that the tetrahedronwhich is above, the third tetrahedron passes above. It is <strong>in</strong>deed because of thatthat we are still here deal<strong>in</strong>g with the Borromean knot.What is annoy<strong>in</strong>g nevertheless, is that even <strong>in</strong> space, even start<strong>in</strong>g from apresupposed spatial, we should also be constra<strong>in</strong>ed <strong>in</strong> this case here to support –s<strong>in</strong>ce when all is said and done, it is we who support it – to support the flatten<strong>in</strong>gout. Even start<strong>in</strong>g from a spatial presupposition, we are forced to support thisflatten<strong>in</strong>g out, very precisely <strong>in</strong> the form of someth<strong>in</strong>g which presents itself as asphere (Fig. V-5b). But what does that mean, if not, that even when wemanipulate space, we have never seen anyth<strong>in</strong>g but surfaces, surfaces no doubtwhich are not banal surfaces because we articulate them as flattened out. Fromthat moment on, it is manifest on the balls that the fundamental plait, the onethat crisscrosses itself 12 times, it is manifest that this fundamental plait formspart of a torus. Exactly this torus that we can materialise by the follow<strong>in</strong>g,namely, the twelve-fold plait, and that we can also moreover materialise <strong>in</strong> termsof the follow<strong>in</strong>g namely, the six-fold plait [Fig. V-3 and Fig. V-4].In truth this function of torus is clearly manifest <strong>in</strong> the balls that I have just givenyou, because it is no less true that between the two little triangles, if we make – Iwould ask you to consider these balls – if we make a polar thread pass through,we will have exactly <strong>in</strong> the same way a torus; for it is enough to make one hole atthe level of these two little triangles to constitute at the same time a torus. This<strong>in</strong>deed is why the situation is homogenous, <strong>in</strong> the case of the Borromean knot, asI have drawn it here, is homogenous between the Borromean knot and thetetrahedron.There is therefore someth<strong>in</strong>g which ensures that it is no less true for atetrahedron that the function of the torus governs here whatever is nodal <strong>in</strong> theBorromean knot. It is a fact, and it is a fact that has strictly never been glimpsed,52

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!