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Projections and tricks tools to test petrogenetic ipotheses Pietro Armienti (Università di Pisa) based on the papers Do We Really Need Mantle Components to Define Mantle Composition? Armienti and Gasperini - J. Petrology 48: 693 and Three-dimensional representation of geochemical data from a multidimensional compositional space. Armienti and Longo - Int, Jour. of Geosciences. In press

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This analysis of data is based on an algorithm which projects data from a multi-dimensional space in three-dimension

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y x A B E E* B A = B + 1 v 1 v 1 v 1 = A - B The projection scheme from R n to R 3 may be envisaged by looking at the analogue for the projection from a plane onto a segment AB. In a 2D space two points B, A visualize the 1D space onto which to map E (x, y) onto E* ( ).

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In a similar way four points materialize the 3D space onto which to project data from R n For simplicitys sake, the four points (end-members) are mapped onto the vertices of a regular tetrahedron

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By choosing a suitable set of four end- members, each defined by its components (e.g. minerals and major elements), the tetrahedron vertices attract the projection of the given analyses. The method allows to project points with negative values of the end- members to obey mass balance constraints

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… allowing also to project data onto a face of the tetrahedron from the opposite vertex. The same scheme allows also to project relevant phase relations in petrologic diagrams

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In systems with many components, compositions are often re-calculated in terms of end members, like in normative calculations that allow to recast the analysis of a rock in terms of fictive anhydrous minerals. SiO 2 TiO 2 Al 2 O 3 Fe 2 O 3 FeOMnOMgOCaONa 2 OK2OK2Op2O5p2O5 p.M.60,085579,88101,96159,69471,84770,93840,30556,0861,979594,1966141,95 Analisi wt%47,721,7117,912,427,610,005,609,953,801,840,45 47,721,7117,912,427,610,005,609,953,801,840,45 47,721,7117,912,427,610,005,609,953,801,840,45

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"canonical" orientation of the tetrahedron whose centre is in the origin of the axes and the X A,Y A,Z A coordinates of vertex A are (0,0,r) while the edge CB forms an angle f with the Y axis (It follows for the X B,Y B, Z B coordinates of B 13) X B = r*sin ( )*sin( ) 14) Y B = r*sin ( )*cos( ) 15)Z B = r*cos( ) being the angle AOB ;

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coordinates of vertices C e D are obtained substituting with ( +120) and ( +240) respectively. Varying causes the tetrahedron to rotate around Z. Rotating around the X axis of an angle, the new coordinates Xi', Yi', Zi' are related to the old ones by the equations : 16) Xi = Xi*cos( )+z*cos( ) 17) Yi = Yi 18) Zi = -Xi*sin( )+Zi*cos( )

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To plot a point P(A p,B p,C p,D p ) within the tetrahedron and get P(Xp,Yp,Zp) we can start projecting E on the edge AB on the basis of the ratio B p /A p, calculating the coordinates of this point P'(X p,Y p,Z p ); 19a) X p = (X A + B p /A p * X B )/(1+ B p /A p ) 19b) Y p = (Y A + B p /A p * Y B )/(1+ B p /A p ) 19c) Z p = (Z A + B p /A p * Z B )/(1+ B p /A p ) A B C D P

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then P is projected onto the face ABC, as a function of the C p /(A p +B p ) and the coordinates of this point P(X p, Y p, Z p ); 20a) Xp = (X p + (C p /(A p +B p )* Xc)/(1+ C p /(A p +B p )) 20b) Yp = (Y p + (C p /(A p +B p )* Yc)/(1+ C p /(A p +B p )) 20c) Zp = (Y p + (C p /(A p +B p )* Zc)/(1+ C p /(A p +B p )) A B C D P P

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At last, the coordinates of P within the tetrahedron are found on the segment PD tracing on PD a segment PP as required by the lever rule : PP/DP= Dp/(Ap+Bp+Cp). 21a) Xp = (Xp +(Dp/(Ap+Bp+Cp).)*XD)/(1+(Dp/(Ap+Bp+Cp)) 21b) Yp = (Yp +(Dp/(Ap+Bp+Cp).)*YD)/(1+(Dp/(Ap+Bp+Cp)) 21c) Zp = (Zp +(Dp/(Ap+Bp+Cp).)*ZD)/(1+(Dp/(Ap+Bp+Cp)) A B C D P P P

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Let us assume that a system E, whose composition is described in terms of n components (e.g. major elements), can be referred to a set of four end members A,B,C,D and that each of them can be represented in terms of the same set of n components. E, A, B, C and D can be represented as points in n-dimensional space (Rn ). E(E1, …En) Considering A, B, C, D as the vertices of a tetrahedron in Rn, we need relations that are able to represent the point E(E1, …En) in n-dimensional space in terms of four end members : A(A1,...,An), B (B1,...,Bn), C (C1,...,Cn), D (D1,...,Dn).

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The algorithm has to assign to E four coordinates (AE,BE,CE,DE) that allow its projection in a 3D tetrahedron that can be easily plotted

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The algorithm is derived extending in Rn the rules for the projection of a point E(e 1, …e n ) in a R 3 tetrahedron: the procedure can be summarized as below: 1.Assign in Rn the coordinates of the tetrahedron vertices: A(a 1,...,a n ), B(b 1,...,b n ), C(c 1,...,c n ), D(d 1,...,d n )

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2 To assign the coordinate AE find in Rn the intersection E' (E'1,...,E'n ) between the line through AE and the hyperplane through BCD, 3 Compute the lengths of segments AE', EE' and AE E

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4. following the lever rule, assign to AE the value : 100-100*(AE-EE)/AE= 100*EE'/AE'. Repeat the steps from 2 to 4 to compute BE,CE,DE changing the vertex and the plane onto which to project E.

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Computing tetrahedral coordinates in this way ensures that for all the points falling inside the tetrahedron AE + BE + CE + DE = 100,

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This sum is different from 100 for points that in Rn lay outside the tetrahedron. To allow the representation of these points in R3 we have to allow for negative tetrahedral coordinates: this is easily accomplished by comparison of lengths of segments AE, EE and AE, it is easy to realize that:

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For points that are inside the tetrahedron AE=AE+EE. For points that are below the face BCD, EE may be smaller or larger than AE, but AE<AE and EE is to be taken in equation 1 2 with a negative sign. For points that are above vertex A, EE > AE and EE has to be taken in equation 1 2 with a positive sign,

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To obtain the 3D projection E* of E: 1) assign tetrahedron vertices (end members) A,B,C,D R n, and compute vectors v 1 =A-D, v 2 = B-D, v 3 =C-D Set and ; w 1 = v 1 and Lastly, set and Tetra: to plot data in rotating tetrahedral diagrams The actual visualization requires a last mapping, namely that from R 3 on the graphical plane R 2, i.e. the computer screen. All the well known drawing maps as Monge orthogonal projections, assonometric projections, perspective may work fine. We chose the vertical (orthogonal) projection on the plane, with a software allowing to choose the size and the orientation of the tetrahedron. Even better, animated rotations do provide a fully satisfactory 3D perception of the spatial distribution of data. Rotation is easily accomplished by using spherical coordinated, starting from a "canonical" orientation of the tetrahedron: set the centre in the origin of the axes and let the coordinates of vertex T 1 be (0,0,r), while the edge T 2 T 3 forms an angle with the Y axis (fig.2). It follows that the coordinates of B are X T2 = r*sin ( )*sin( ) Y T2 = r*sin ( )*cos( )Z T2 = r*cos( ) where is the angle T 1 O T 2 ; the coordinates of vertices T 3 e T 4 can be obtained from the above equations substituting with ( +2 /3) and ( +4 /3), respectively. Varying causes the tetrahedron to rotate around Z axis. If tetrahedron is also allowed to rotate around the Y axis of an angle, the new coordinates X i ', Y i ', Z i ' of each vertex are related to the old ones by the equations : X i = X i *cos( )+z*cos( ) Y i = Y i Z i = -X i *sin( )+Z i *cos( ) The above equations allow dependence of vertices (T 1,T 2,T 3,T 4 ) from and and the relation E** = 1 T 1 + 2 T 2 + 3 T 3 + 4 T 4 Maps the same tetrahedral coordinates in R n with respect to ABCD on a rotated tetrahedron in R 3, to visualize the data set from different points of view. Step by step increments of and animate rotations revealing possible clusters or special arrangements on planes or lines. All the figures of this paper can be downloaded as Power Point animations from the site.http://www.dst.unipi.it/dst/armienti/download.htm. Animations exploit the potentialities of tetrahedral diagrams at its best, showing 3D relations among data arrays. Observe that the adopted procedure maps the points of the perhaps highly irregular tetrahedron ABCD onto those of a regular one T 1,T 2,T 3,T 4. This provideshttp://www.dst.unipi.it/dst/armienti/download.htm

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2) For each point to project compute the vector E=E- D and its orthogonal projection E on, that is E = 1 w 1 + 2 w 2 + 3 w 3 where i =, i = 1, 2, 3. E is the linear combination of w 1,w 2,w 3 nearest to E in R n. Since the distance is translation-invariant, it follows that E*=D+ 1 v 1 + 2 v 2 + 3 v 3 is the point in the 3D-space passing through A,B,C, D nearest to D + E=E.

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3) Finally, remark that E*= D + 1 w 1 + 2 w 2 + 3 w 3 = D+ 1 (A–D)+ 2 [ B–D - (A–D)] + + 3 { C–D- (A–D) - [ B–D- (A–D)]}= =( 1 - 2 - 3 + 3 A ( 2 - 3 B 3 C + + ( 1 - 2 - 3 + 2 + 3 + 3 - 3 D The sum of the four coefficients of the end members at the right hand of the above formula, is 1; therefore, they represent the tetrahedral coordinates i of E* with respect to the four vertices A,B,C, D:

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Tetrahedral coordinates 1 = 1 - 2 - 3 + 3 2 = 2 - 3 3 = 3 4 = 1 - 2 - 3 + 2 + 3 + 3 - 3 1 - 2 - 3 The point E* coincides with E if and only if E itself is a linear combination in R n of the end members with sum of coefficients equal to one. Otherwise, E* is distinct from E, while enjoying the property to be at the minimal Euclidean distance from it, among all points of the (affine) subspace passing through the end members.

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Let now (T 1,T 2,T 3,T 4 ) be the vectors in R 3 to which the tetrahedron vertices A,B,C,D are mapped on ; Finally, E will be mapped in R 3 onto E* by the relation E*= 1 T 1 + 2 T 2 + 3 T 3 + 4 T 4. For a regular tetrahedron a possible choice is : T 1 =(0,0,1), T 2 =(2* 2/3,0,-1/3); T 3 =(- 2/3, 2/ 3, -1/3); T 4 =(- 2/3, - 2/ 3, -1/3).

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