Projection tensor in from (m+n) dim down on n-dim

In summary, we can construct a projection tensor onto a subspace of a tangent space using the coordinate basis and its metric duals, and this projection tensor satisfies certain properties.
  • #1
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Suppose that we have an (n+m)-dimensional tangent space ##T_p^{n+m}## which we decompose into the direct sum of two tangent spaces ##T_p^{n+m} = T_p^n \oplus T_p^m##. We have a coordinate basis in some region of the manifold ##\left\{\partial_{\mu}\right\}_{\mu=1}^{n+m}## from which we want to construct an adopted basis ##\left\{e_a, e_k\right\}## such that ##e_a \in T_p^{n}## and ##e_k \in T_p^m## with ##a = 1,2,\ldots , n## and ##k = 1, 2, \ldots m##. Now if ##m=1## one can easily construct a projection operator which projects down on the ##n##-dimensional tangent-space ##T_p^n## by

$$P = \mathbb{1} - \frac{e_{n+1}\otimes e^{n+1}}{g_{(n+1)(n+1)}} $$

where I have defined ##e_k \equiv e_{n+1}## and the metric-dual ##e^k = g(e_k, \ )##, where ##g## is the metric tensor and ##\mathbb{1}## is the identity tensor. As a projection tensor must, we have that ##P## satisfies

$$P P = \mathbb{1} - 2 \frac{e_{n+1}\otimes e^{n+1}}{g_{(n+1)(n+1)}} + \frac{1}{g^2_{(n+1)(n+1)}} e_{n+1} \otimes e^{n+1} e^{n+1}(e_{n+1}) = \mathbb{1} - \frac{e_{n+1}\otimes e^{n+1}}{g_{(n+1)(n+1)}} = P.$$

On the other hand, when ##m > 1## the similar expression

##P = \mathbb{1} - \sum_{k=1}^m\frac{e_{k}\otimes e^{k}}{g_{kk}}##

does not automatically satisfy ##PP = P##, since

$$PP = \mathbb{1} - 2 \sum_{k=1}^m\frac{e_{k}\otimes e^{k}}{g_{kk}} + \sum_k \sum_l \frac{g_{kl}}{g_{kk}g_{ll}}e_k \otimes e^l. $$

So it's only a projection tensor if ##g_{kl} = g_{kk} \delta_{kl}## no, sum. Is there a general formula for the projection tensor down on ##T_p^n## that satisfies the required relations and for which we need not have ##g_{kl} = g_{kk}\delta_{kl}##?
 
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  • #2


Hello,

Thank you for your interesting question. In general, the projection tensor onto a subspace of a tangent space can be written as

$$P = \sum_{i=1}^n \frac{e_i \otimes e^i}{g_{ii}},$$

where ##e_i## are the basis vectors of the subspace and ##e^i## are the corresponding metric duals. This projection tensor satisfies the properties ##P^2 = P## and ##P(e_i) = e_i##, as required for a projection tensor.

In the case where ##m > 1##, the projection onto the subspace ##T_p^n## would be given by

$$P = \sum_{i=1}^n \frac{e_i \otimes e^i}{g_{ii}} + \sum_{j=n+1}^{n+m} \frac{e_j \otimes e^j}{g_{jj}},$$

where ##e_j## are the basis vectors of the remaining ##m## dimensions. This projection tensor would also satisfy the properties of a projection tensor.

I hope this helps. Let me know if you have any further questions.
 

1. What is a projection tensor?

A projection tensor is a mathematical tool used in linear algebra and geometry to project a higher dimensional space onto a lower dimensional space. It essentially maps points from a higher dimensional space onto a subspace.

2. How is a projection tensor represented?

A projection tensor is typically represented by a square matrix with elements that determine the projection onto the subspace. It can also be represented using tensor notation, which is a more compact way of writing linear algebra equations.

3. What is the purpose of a projection tensor?

The purpose of a projection tensor is to transform a higher dimensional space into a lower dimensional space while preserving certain properties, such as orthogonality or distance. This can be useful in various fields, such as computer graphics, physics, and machine learning.

4. How is a projection tensor calculated?

The calculation of a projection tensor depends on the specific problem it is being used for. In general, it involves finding a basis for the subspace onto which the projection will be made, and then using this basis to construct the projection matrix. This can be done using techniques such as Gram-Schmidt orthogonalization or singular value decomposition.

5. Can a projection tensor be used for non-linear projections?

No, a projection tensor can only be used for linear projections. Non-linear projections require more complex mathematical tools, such as tensors with higher order components.

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