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Our more modest goal for this article is to show how to enter covariant and contravariant tensors, compute their covariant derivatives, obtain the equations of parallel transport and geodesics, and compute the basic tensors of general relativity. Declaring the Frame in DifferentialGeometry.

Before a vector or tensor can be entered, a frame must be declared by stating its variables and giving it a name. For example, to declare as the Cartesian space with variables and , execute. The default behavior for DifferentialGeometry is that from this point onward, the prompt would be modified to display the frame name for as long as that frame were the active one. Thus, without our having suspended this default with the Preferences command in the Initializations section, the prompt following the DGsetup command would be the one shown in Table 2.

We have elected to suspend this default behavior for three reasons. First, interactive editing of a worksheet can create a confusing display. If in a section where one frame name appears in all prompts, a new frame is defined, the subsequent prompts that show the earlier name do not change to show the new name.

Only a "new" prompt will display the new frame name. The result is a worksheet with prompts showing different frame names where they might not be relevant.

Such misplaced prompts have to be deleted manually if they are not to provide incorrect information. Second, these modified prompts are persistent - they cannot be removed by any Maple command. They have to be removed by deletion. The Maple interface command that modifies prompts does not cascade the change through existing prompts.

It only modifies new prompts.

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Finally, if the commands are executed in Document Blocks, and not at prompts, there will be no visible prompt to modify. Thus, the modified prompt is a worksheet paradigm that does not carry through all of Maple usage. For all these reasons, we will not have frame names visible in our prompts. Anyone who had taken even a perfunctory dip into the waters of tensor calculus knows there are two words, covariant and contravariant , that must be faced.

We will not be able to enter a tensor in the Tensor package without making the distinction between these two terms.


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Using the Einstein summation convention repeated indices, one raised and one lowered, are summed , Table 3 defines contravariant and covariant vectors. Vector Type. In the rightmost column of Table 3 uses the notation and for components expressed in the -coordinate system, but and for components in the -coordinate system. Texts also denote the new coordinate system bu the use of an overbar on the component, or a prime on the left side of the variable. The components and are the contravariant and covariant components, respectively, of the vector V. The basis vectors and are reciprocal, so that.

Thus, an orthonormal basis is self-reciprocal. That is why the distinction between contravariant and covariant basis does not matter in Cartesian spaces. If is the mapping from to via functions of the form , then the gradient vectors are the rows of the Jacobian matrix , where the upper index is interpreted as a row index, and the lower index , as a column index. If is the mapping from to via functions of the form , then the tangent vectors are the columns of the Jacobian matrix. To facilitate the implementation of the contravariant transformation law, writing the components as a column vector v means the sums with the Jacobian matrix are along a row and across the columns of the matrix.


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Hence, the matrix product implements the contravariant transformation. Writing the components as the row vector w means the sums with the Jacobian matrix are down a column but across the rows.

Multilinear Functions of Direction and Their Uses in Differential Geometry (Paperback)

Hence, the matrix product implements the covariant transformation. This inherent distinction between tangent bases and normal bases induces the distinction between contravariant and covariant. Using column and row vectors to express this difference is a convenient visual device in classical tensor calculus. If is a vector space, say, with basis , then a rank-two tensor is a multilinear object from , the direct product of with itself, having doubly-indexed basis objects. The tensor is actually the object. Of course, the are the contravariant components of the tensor; and just as for vectors, there would be the equivalent covariant components,.

There are even mixed tensors that transform contravariantly in one index but covariantly in another. In actual practice, one manipulates just the components of the tensor, and almost never explicitly exhibits the basis objects. However, in the DifferentialGeometry package, vectors and tensors require an explicit use of the basis objects.

The basis for could be entered as. Alternatively, enter such expressions in text linear, 1D mode and convert to math mode via the Context Menu. The reciprocal or dual basis is then. In actual fact, is considered a differential form , more in keeping with the modern approach to differential geometry. Representing Vectors as DifferentialGeometry Objects. In the DifferentialGeometry package, the contravariant vector whose components are is given by.

The evalDG and the DGzip commands are two of the simpler ways to create an object whose data structure is intrinsic to the DifferentialGeometry package. When using the evalDG command, the asterisk is the explicit multiplication operator. If this vector had been entered in math mode, the echo of the asterisk would be a centered dot.

The alternative to the asterisk would be the space. The covariant vector whose components are is entered as. There does not seem to be a simple way to represent a vector as a column or row vector. The Tools subpackage provides the DGinfo command with which the components of a vector can be extracted. Its use is illustrated by. Representing Tensors as DifferentialGeometry Objects. The construct corresponds to the dyadic basis element , etc. The components of a tensor can be recovered with the DGinfo command. The components of the rank-two tensor are often represented as the entries of a matrix.

If a matrix is used for such a representation, it is possible to convert the matrix to a DifferentialGeometry tensor. Given two tensors of conformable dimensions, say and , forming the sum of products is the operation contraction of indices. Recall that the Einstein summation convention indicates a sum on the same index when it appears once raised and once lowered.

This operation, implemented in the Tensor package with the command ContractIndices , is illustrated for the two tensors.

Multilinear Functions of Direction: And Their Uses in Differential Geometry by Eric Harold Neville

To simplify data entry, we rename the first as T and the second, as V. The four possible contractions that result in a rank-two tensor are given in Table 4, where their components are displayed as elements of matrices. Careful inspection of the four tensors in Table 4 see especially the rightmost column shows that they are all different.

The geometry of a manifold is first captured in the covariant metric tensor or its contravariant counterpart. There is no "calculus" in tensor calculus without first obtaining this essential tensor. Hence, it is imperative that there be efficient ways to obtain this tensor. Several of these techniques will be illustrated for the Cartesian plane on which polar coordinates have been imposed. Method 1 - Obtain as a Matrix and Convert to a Tensor. Define the map with equations of the form via. Then, a representation of the basis vectors is given by. To convert this to the metric tensor in polar coordinates, we need to define the polar frame with.

The covariant form of this metric tensor, , can be obtained with. The matrix representing is the inverse of the matrix representing. Begin with the contravariant metric tensor on the Cartesian space :. The same vector V can be given with respect to the natural tangent basis vectors or with respect to the reciprocal basis of gradient basis vectors. Thus, one has. The conversion between contravariant and covariant components of V is effected by contraction with the metric tensor:. Thus, we have. Alternatively, given the covariant vector. Once the metric tensor is known, the way the basis vectors change from point to point can be determined.

It turns out that the rate of change of the basis vectors can be expressed as linear combinations of these same vectors. The coefficients of these linear combinations are called the connection coefficients , or the Christoffel symbols. Depending on the form used for these symbols, they are called the Christoffel symbols of the first kind or the second kind.

Since there is a built-in command that provides the Christoffel symbols of the second kind, we will obtain those via.

Classical texts in differential geometry use the notation , , or for Christoffel symbols of the second kind, and or for Christoffel symbols of the first kind. Consequently, the correct interpretation of the output of the Christoffel command is captured, painstakingly and laboriously, in Table 5. The closest the Tensor package comes to articulating the Christoffel symbols is. Thus, the middle index in the list is the raised one, and the first and third are the lower ones.

For polar coordinates, show that. Thus, show that the derivative of a basis vector can be expressed as a linear combination of the basis vectors, and that the coefficients of these linear combinations are the Christoffel symbols. To show that derivatives of basis vectors can be expressed as linear combinations of basis vectors, begin by expressing in terms of.

This is done in Table 6 where the symbols i and j are introduced explicitly and naming clashes are avoided by introducing the alternate names for. Table 7, which lists the derivatives of the basis vectors, shows that the Christoffel symbols indeed are the coefficients in the linear combinations of basis vectors that express the derivatives of the basis vectors. Table 8 provides a formula for computing Christoffel symbols of the second kind from the components of the metric tensor. A practical notational shortcut is the use of for the differentiation operator.

The formula in Table 8 would be easier to write and even remember if this notational device is used. In calculus, is the directional derivative of the scalar function taken in the direction of the unit vector u. The gradient vector arises naturally from the calculation of the derivative. Of course, this calculation extends to higher dimensions. But more important, note how starting with a scalar function , the vector quantity must be defined, and the desired directional derivative is a dot product of this gradient vector with the direction vector.

The covariant derivative arises in much the same way, that is, from defining a directional derivative of a vector. That is, it is a different multilinear map on different tangent planes. I realize that on every point it is a different function. But if my analogy to that of vector fields is correct, then these two formulations should be equivalent. Or am I missing something again? Thank you for your reply.

I guess the description by Lavinia corresponds to the construction mentioned by jgens. Thank you again ;. Homework Helper. However, one can still use this simplifying incorrect version, if one has an embedded manifold in euclidean space. Here the group operation again allows us to translate tangent vectors to the origin. Last edited: Nov 28, Then again, we did not really prove anything with the tangent space, just some intuitive explanations about orientation which we rigorously defined using the jacobian of the transition functions.

So, I guess Spivak's approach is still better ;. Spivak is Spivak after all.. I'd also like to use this chance to thank you Mathwonk for all the high quality content you have been posting for so many years. You are one of the main reasons I've been lurking around these forums for several years now; in fact during the period you stopped posting, I also stopped visiting PF. Related Threads for: Confusion regarding differential forms and tangent space Spivak,Calc. Posted May 9, Replies 4 Views 3K.

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