Introduction to algebraic geometry pdf

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Create a free Team What is Teams? Learn more. Asked 11 years, 5 months ago. Active 2 years, 8 months ago. Viewed 26k times. Unless it's the case that asking for undergraduate textbooks especially would result in different answers.

Add a comment. Active Oldest Votes. I have a question. And now im about to join a class in algebraic geometry. Being away for so long i have forgotten much of the basics and i wonder if you possibly can recommend some books to read before i dive into algebraic geometry. Books about subjects you need to know to understand algebraic geometry. I have forgotten things like rings, fields, most of complex analysis etc. An introductory book about algebraic structures maybe is a good start?

Anything else? Thank you. Follow the first links in this other answer of mine math. Once "basic algebra" is mastered, any good comprehensive reference for adv. Show 5 more comments.

Georges Elencwajg. Note that ShreevatsaR's sentence should be parsed as, "with little [math] background, and with interest in CS," as opposed to, "with little background and interest in CS "! Please feel free to edit the answer so that it's more appropriate.

Such interpretation allows for applying directly most of the results obtained above. In particular, if A is invertible, one may transform the augmented matrix in such a way that the identity matrix is on the left.

Then, on the right one has the unique solution, i. Linear Homogeneous Equations We consider a particular type of matrix equations or systems of linear equations with a vanishing righthand side, referred to as homogeneous.

In this setting our work is greatly simplified, e. One more convenience provided by the homogeneous case is that we do not need to bother about the extensions when applying the Gauss algorithm as it simply does not affect them.

Another property of linear homogeneous systems is given by the following Theorem 9 The solutions of a linear homogeneous matrix equation and in particular, system of linear equations constitute a vector space. Suppose also that A has maximal rank, i. Note the two identity matrices belong to vector spaces of different dimension, so their equality is possible and unavoidable only in the case of square matrices.

Note, however, that the Kronecker-Capelli theorem still applies to this case and the rank of A B may exceed that of A, since we add more columns in the extension. Therefore, one needs to check first whether the two ranks agree. On the other hand, the familiar Gauss method works here with almost no modifications. Then, the remaining rows will either be trivial, in which case the system is simply over-determined, or contain a contradiction and thus, yield inconsistency.

Let us illustrate this with several examples. Provide an example, in which the inequality does not hold. What is the explicit relation between the two?

Chapter 8 Vectors in R3 8. For example, linear combinations are the most fundamental operations in any vector space and the notion of linear dependence remains identical, i. Moreover, the number of independent vectors in a given system usually re- ferred to as its rank cannot exceed the dimension n of the corresponding space. Now, consider the expansion 8. This crucial difference between upper and lower indices remains hid- den in the case of orthonormal bases, in which the two types of coordinates coincide and the transition matrix T is clearly equal to its contragradient.

Such transformations are called orthogonal and they play a central role in both geometry and mechanics. As it turns out, the positive sign of the determinant corresponds to an orientation-preserving transformation, which in our case reduces to a pure rotation.

The product is obviously skew-symmetric by construction why? Note that in each case the double cross product lies in the plane determined by the two vectors in the brackets explain why and prove the equalities!

There is one more generalization of the wedge product in R2. This time we put the emphasis on the fact that it yields the oriented area determined by its two factors. One consequence is that coplanar vectors have vanishing triple product. However, we are going to need normalization factors here too, and it is not difficult to see that the triple product, i. Note also that the concept of duality is always mutual, i.

The dual basis technique naturally extends to higher dimensions as well, but then one does not have the convenience of using the cross product. Try to formulate an analogous construction for Rn based on the exterior product. Prove the particular case of formula 8. Find also the coordinates of the geometric center, the total surface area and the distance from the point D to the plane determined by A, B and C.

Did you have fun? How about a two-parameter set? Try to explain this coincidence algebraically. However, one should be careful, since there are differences too. To begin with, we may still consider two types of vectors - attached and free, typically represented by radius-vectors of points and their differences.

The latter constitute the vector space R3 , while the former, the affine space R3 as a space of points. Thus, one may think that there are two copies of R3 involved in our geometric considerations. When lines and planes come into play, more complicated spaces appear naturally in the model, such as Grassmannian and projective spaces, but we are not going to discuss them here explicitly.

For example, the vector parametric equation of a line in R3 is written in the exact same way as in R2 See Figure 9. Note, however, that here, unlike in the two-dimensional case we have two, rather than one linear equation. The above is actually referred to as the general equation of a plane and its geometric interpretation is quite similar to that of a general equation of a line in R2 See Figure 9. Similarly, if we generalize the intercept-intercept equation of a line 5.

Things remain relatively simple as long as there are only points and vectors to consider. Then, we also have incidence relations, e. As for the parametric case 9. Let us now consider the mutual position of two lines: g, determined by its direction vector t and a point A, and h - by the pair s and B, respectively.

Similarly, two non-parallel planes intersect at a line, which is the one-parameter solu- tion to the system of two equations for three variable. In particular, this distance may be zero if the two planes coincide or the line belongs to the plane, respectively.

Finally, let us consider the case of three planes in R3 from the perspective of systems of linear equations and the Kronecker-Capelli theorem in partic- ular. In the non-degenerate case, in which the system of the three normals has maximal rank, the planes are bound to intersect at a single point much like the coordinate planes OXY , OY Z and OZX intersect at the origin. The other extreme setting involves parallelism of two or even three normals corresponding to rank two and rank one, respectively.

In between, how- ever, we see something with no analogue in R2 : neither two of the normals are parallel and yet, the system is either dependant of inconsistent, i. Consider other possible settings. Now, the rank of the system the above vectors constitute determines whether the four points are coplanar or in a generic position. Another typical task in stereometry is to determine the mutual position and distance between two lines.

Let us take for example the one passing through the pair of points A and B we shall denote it with g and another one say, h containing C and D. See if it yields the same result. For a better understanding of the geometric properties of operators it is convenient to introduce the notions of range and kernel. Obviously, due to linearity, ker A always contains the zero in V and is said to be trivial if it does not contain anything else.

In this case A is called 1 projecting spacial vectors onto the XY -plane we lose track of the vertical component. A non-trivial example would be a reflection or a rotation in the plane - it covers all R2 and the image of each non-zero vectors is also a non-zero vector.

More generally, as we shall see later, each square matrix with non-vanishing determinant may be thought of as a matrix of a linear isomorphism. Proof: We only give a sketch of the proof here. In the next section we provide more explicit geometric examples. Since this course is focused mainly on Euclidean geometry, we shall pay special atten- tion to operators related to planar and spacial motions. In order to do this we only need to note that the parallel to u is preserved by Mu , while the one in the normal direction changes its sign, i.

On the other hand, spacial reflections are usually taken with respect to a plane, or some more general mirror surface.

Actually, our construction can be used directly only for free vectors, not for radius-vectors explain why and we already know a straightforward way to obtain them in terms of coordinate differences. There is another, rather physical, setting that involves reflection with respect to a vector, rather than a plane.

In order to obtain the trajectory of a particular reflected ray or elastically scattered particle, e. In physical considerations one may introduce a time parameter t and divide There are more examples of this type at the end of the present chapter. Rotations and the Galilean Group As we already mentioned, all spacial motions are generated by reflections. Since translations are reduced to vector summation, i. These properties turn the set of planar rotations into a commutative matrix group, i.

Adding translations to that, which is allowing each radius-vector to be shifted arbitrarily, one obtains the group of Euclidean motions in the plane. Note, however, that the translational and rotational components of such motions are intertwined and hence, do not commute.

The block-matrix technique works similarly in every dimension and the exten- sion of the translational counterpart is trivial, so the main difference is the group of spacial rotations compared to the planar case. To begin with, it is non-commutative, which makes it far more difficult to study. Each spacial rotation is again rotation in a plane or about the axis, perpendicular to that plane , which cannot be said for higher dimensions. This plane, however, is not fixed but varies within the group.

In particular, rotations about the coordinate axes adopt a simple form, e. Baring in mind the above considerations about the rotation group in R3 , one may still express spacial motions in pretty much the same block-matrix form as in R2 , i. Note that even if the rotation group is commutative, which is only in R2 , introducing translations to it certainly destroys this property as the above composition formula clearly indicates.

This is because translations and ro- tations are entangled in the Galilean motion group in a structure referred 3 clearly, the cross product is responsible for the non-commutativity and when the two axes are collinear the expression is reduced to an addition formula for the tangent function.

This peculiar behavior may be encoded in a conveniently chosen algebraic struc- ture, which naturally adopts similar properties. In mechanics this is also known as screw calculus. Similar algebraic con- structions often make the description of cumbersome geometric and phys- ical theories way easier and more productive, e. Complex numbers, quaternions and groups themselves provide a good illustration of this idea.

Determine the rank and the nullity of a linear operator that projects n-dimensional vectors onto a k-dimensional subspace of Rn. Given the four points A 1; 2; 3 , B 2; 3; 1 , C 3; 1; 2 and D 0; 0; 0 , find the coordinates of the reflections of the geometric center of the pyramid ABCD with respect to its four sides.

Find the trajectory of the ball after the reflection. This is especially useful in me- chanical considerations including optimization, determining states of equi- libria for complex systems etc. However, we are interested only in non- trivial solutions, which we refer to as eigenvectors.

In particular, the construction of of eigenvectors and eigenval- ues turns out to be of crucial importance for the study of skew- symmetric, orthogonal and unitary operators with their geometric and dynamical prop- erties. For the description of the nontrivial solutions to equation With the aid of condition For a simple root the corresponding solution contains one undetermined pa- rameter, which corresponds to the freedom of scaling v. We shall clarify this part by the end of the chapter.

We begin with a simple but powerful Theorem 12 The eigenvalues of symmetric and hermitian operators are all real and the eigenvectors corresponding to different eigenvalues - orthogonal. If A is hermitian, we also have from the conjugated version of formula The symmetric case follows as a subset.

In the above cases the eigenvectors constitute an orthonormal basis1 , in which the matrix of the corresponding linear operator has a diagonal form. The orthogonal case obviously follows as a subset. Let f be a smooth function. Therefore, the matrices A and f A commute, so they determine identical sets of eigenvectors prove this point alone! Furthermore, for non-smooth functions f , one may always find an approxi- mations with smooth ones and all properties we need endure the limit.

We shall leave the details of the proof to the reader providing only an motivational example. Now, if two matrices are diagonal in the same basis, they obviously commute and vice versa - if one of them is diagonal in a given basis and the commutator vanishes, the other needs to be diagonal as well if this statement is not obvious to you, work out the details.

Our main example is the exponential function, which maps skew-symmetric skew-hermitian matrices into orthogonal unitary ones. This is also one more way to see that the eigenvalues of the latter lie on the unit circle since for the former they are purely imaginary.

Moreover, we may prove that each rotation, i. No axis in R4 is preserved by its action, only the origin the rotation center remains fixed. In every other case we may encounter rotation in a plane at best, but this is something that does not happen often as it demands a set of highly non-trivial conditions to be satisfied by the corresponding matrix entries.

Similarly, we have invariant planes and higher-dimensional subspaces in Rn. How- ever, since the whole plane spanned by v1 and v2 which, by the way, is the orthogonal complement of v3 is preserved by the action of A, we may choose an arbitrary pair of vectors in this plane and in particular, an orthonormal one. Similarly, if A is a matrix of a bilinear form, i. If we switch from one orthonormal basis to another, as we do in our example, the matrix T is orthogonal and the two transformations would be identical, but this is not so in the generic case.

Of course, not. In this case, the matrix we are dealing with is neither symmetric, nor skew- symmetric and its eigenvectors do not necessarily constitute a basis in R3. Actually, it is not difficult to see that its rank is equal to two: there is no pair of linearly dependent rows or columns, so we have only one relation for the three vectors. The linear operator B acts on vectors in R3 as a cross product with v followed by a sign inversion of the first two components.

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