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8

If $V$ is a finite-dimensional vector space then there exists an isomorphism $V\cong V^*$, under which every vector in $V$ can be interpreted as a covector (linear map $V\to $ the scalar field). However, there are going to be many such maps, and there is in general no canonical such isomorphism, so given a vector $v\in V$, which covector in $V^*$ it ...


6

To add to all the answers above, there is a delightful example in the text "Mathematics for Physics" by Stone and Goldbart, (Appendix A.3), to clarify the difference between vectors and co-vectors, which I can't resist quoting here. One way of driving home the distinction between $V$ and $V^*$ is to consider the space $V$ of fruit orders at a grocers. ...


3

Expanding just a touch on my comment: One simple way to see that vectors and covectors are not as naively identifiable as your friend says is to realize that there are settings where they aren't even isomorphic. They are developed independently on abstract manifolds, via the tangent and cotangent structures, respectively. In particular, the only reason ...


3

1), 2): yes. On Manifolds you can introduce a metric, which makes the manifold a Riemmannian manifold (assuming the metric is positive definite). A metric, by definition, is a (0,2) tensor field which defines a scalar product on the tangent bundle, i.e. the metric in a point $p$ is a scalar product on the tangent space at $p$. This, in local coordinates, has ...


2

In Lee's 'Intro to Smooth Manifolds', $\Lambda^k(V)$ refers to the space of alternating $k$-tensors on a vector space $V$, as you mentioned. However, the space $\Omega^k(M)$ is the space of smooth $k$-forms on a smooth manifold $M$. That is, an element of $\omega \in \Omega^k(M)$ is a smooth map $M \to \Lambda^k(T^* M)$ (called a smooth section of ...


2

It always exists (at least, for finite-dimensional spaces), after weakening positive-definite to positive-semidefinite, regardless of whether $A$ is invertible; it can be constructed from the singular value decomposition, for example. The decomposition is not unique if $A$ is singular, though; take the zero matrix for a trivial example.


2

Actually, when you write $(A \times B)_i$, this is no longer a vector, but just one of its components. The vector itself is $$(A \times B) = ((A \times B)_1, (A \times B)_2, (A \times B)_3)$$ Having this in mind, the expression $$(A \times B)_i = \epsilon_{ijk}A_j B_k$$ means three equations, for $i$ ranges from $1$ to three. The equations are: $$(A \times ...


2

This is simple and you are able to prove it on your own. Write down the definitions on both sides and compare what the result is. Hint: You have to check two different cases.


1

Note that for general curvilinear coordinates, the property $$ \delta_{ij} = 0, \: i \neq j \quad \delta_{ij}= 1, \: i=j $$ even if true in one basis, is NOT true in general, because $\delta_{ij}$ transforms as a (0,2) tensor. However, the (mixed) (1,1) Kronecker delta $\delta^i_j$ does have this property.


1

We have (where $(\cdot, \cdot)$ denotes the inner product) \begin{align*} Sv &= \sum_i (Sv, e_i)e_i\\ &= \sum_i \left(\sum_j v_je'_j, e_j\right)e_i\\ &= \sum_i \left(\sum_j v_j(e'_j, e_j)\right)e_i\\ &= \sum_i \left(\sum_j v_j\beta_{ji}\right)e_i \end{align*} So $(Sv)_i = \beta_{ji}v_j$, that is $R= S$.


1

Hint for part I: Consider how the product $g_{\mu\nu} \; x^\nu$ transforms, since you know how the individual terms in the product transform under a change of basis. You will see that it transforms exactly like $x_\mu$. For the second part of your question: "a vector that is neither contravariant or covariant" is not a vector (in the sense used in ...


1

This is somewhat late in the day / year, but I suspect the author is asking about representations of isotropic Cartesian tensors, where "isotropic" means "invariant under the action of proper orthogonal transformations" and "Cartesian" means the underlying space is Euclidean $R^n$ ($R^3$ is a case of great practical interest). The proofs for the two cases ...


1

You've forgotten to multiply $T\cdot\Lambda $ by $T^{-1}$ on the right - when we change bases/coordinates we have to make sure that our linear map is expressing a transformation in this new basis. The $T^{-1}$ above ensures this description is complete. The resulting matrix $T\cdot\Lambda \cdot T^{-1}$is then $similar$ to our original matrix $\Lambda$. As ...


1

I will work in index notation and with the Einstein summation convention. First, let's write the LHS in index notation and simplify: \begin{align} \partial_i \left(B_i B_j -\frac{1}{2} B_k B_k \delta_{i j}\right) =(\partial_i B_i)B_j +B_i (\partial_i B_j)-B_k( \partial_j B_k). \end{align} Here we have applied the product rule and summed over the delta ...


1

As an alternative to Semiclassical, the problem can be tackled entirely using the machinery of clifford algebra. Let $a$ be an arbitrary vector, and let $\underline B(a) = \overline B(a) = (B \cdot a) B$. Then we have $$\nabla \cdot [\underline B(a) - \frac{1}{2} B^2 a] = a \cdot [\overline B(\nabla) - \frac{1}{2} \nabla B^2] = a \cdot [B (\nabla \cdot B) ...


1

I guess I'd write that as $A B^{t}$. But I don't think that gets at the heart of what you're asking. And the following may help or not...but I offer it up anyhow. To really get a grasp of the difference of the two approaches, pick up a classical differential geometry book -- something like O'Neill's Elementary Differential Geometry, or perhaps do Carmo's ...


1

I think you have some misconceptions about what you are talking about. In any given basis any linear operator is represented by a fixed matrix. When we talk about linear operators or tensors we are talking coordinate free, when we talk about matrices then we are not. When you write $a_{ij}b_{kj}$ you already have some coordinates. If $A=a_{ij}$ and $ ...


1

Let's take an orthogonal basis of $\mathbb{R}^3$: $e_1, e_2, e_3$. We can write vectors $a = \sum_i a_i e_i$ and $b = \sum_i b_i e_i$. The oriented subspace spanned by these vectors is a 2-blade, denoted $a \wedge b$: $$a \wedge b = \left(\sum_i a_i e_i\right) \wedge \left(\sum_j b_j e_j\right) = \sum_{ij} a_i b_j e_i \wedge e_j$$ where in the last line we ...



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