What is a tensor with two upper or lower indices?

Question

I know very little about tensors. I am trying understand if the following expression would make $$x^a=L^a_{\;b} M^{bc}\hat{x}_c.$$ Einstein summation convention applies.

So here is what I think I understood:

$x^a$ is the $a$-th component of vector $x$ expressed in some basis $B$.

$\hat{x}_c$ is the $c$-th component of a covector $\hat{x}$ expressed in a basis $B_2$.

$L_b^a$ is the entry in position (a,b) of a matrix that represents a linear mapping $L$ using a basis $B_3$.

So It must be that $M$ takes a covector and outputs a vector. Because $L$ can only take vectors.

So as vectors $x=L(M\hat{x}).$

Can you give me an intuitive understanding of what $M$ can be?

Answer 1

In special relativity, the special case where $L^a_{\;b}$ is a Lorentz transformation allows us to relate two coordinate systems' representation of the same vector, viz. $x^a=L^a_{\;b} x'^b$. This coincides with your equation provided $x'^b=M^{bc}\hat{x}_c$, as happens e.g. with $\hat{x}$ the index-lowered equivalent of $x'$ and $M^{bc}$ the index-raising metric tensor $g^{bc}$, viz. $g^{bc}g_{cd}=\delta^b_d$ with $g_{cd}$ the index-lowering tensor.

We typically adopt the notation $y_d=g_{bd}y^b$. Thanks to the above analysis, we know that if instead $\hat{x}_c=Q_c^{\;d}x'_d$ for some linear $Q$, equating two expressions for $x^a$ gives $L^a_{\;b} x'^b=M^{bc}Q_c^{\;d}x'_d$, i.e. $M^{bc}Q_c^{\;d}=g^{bd}$. In matrix terms $MQ=g^{-1}$, since we typically use the "matrix" $g$ to denote the index-lowering tensor. Thus $M=g^{-1}Q^{-1}$ provided $Q$ is invertible. This expands the set of legal choices for $M$ to all invertible matrices. For that matter, $(LM)^{ac}=L^a_{\;b}M^{bc}$ can be any invertible matrix too, if $L$ is invertible.

Answer 2

Immagine that in the space you're working there exists a scalar product: $$(x, y)=x^aG_{ab}y^b$$ You can see that $G_{ab}$ is the matrix that (multiplicatively) maps vectors to covectors.

Let $$y_a=G_{ab}y^b\tag{1}$$ be the covector corresponding to the vector $y^b$ through $G_{ab}$, that is, the covector that gives the scalar product of $y^b$ times any vector $x^a$ by simply summing up the product of the homologous components of itself ($y_a$) and of $x^a$: $$(x, y)=x^ay_a$$ It results that $$M^{bc}\hat{x}_c=M^{bc}G_{cd}\hat{x}^d=M^b_d\hat{x}^d$$ where $M^b_d$ is a linear mapping (of vectors into vectors).

So given a scalar product (that is, $G_{ab}$), $M^{bc}$ represents numerically just the same linear mapping $M$ represented by $M^b_d$ but, instead of operating directly on vectors as $M^b_d$ does, it operates on those covectors which correspond to vectors by $(1)$, that is $$M^{bc}=(G^{-1})^{cd}M^b_d$$

Answer 3

If the tangent space at point $p$ in a real manifold $M$ is $T_p$ and the cotangent space is $T_p^*$ then a tensor like $L^a_b$ with one upper index and one lower index represents a linear map from $T_p \times T_p^*$ to $\mathbb{R}$. So you can think of $L^a_b$ as a linear operator that consumes a vector with co-ordinates $v^b$ and a covector with co-ordinates $u_a$ and outputs a scalar $L^a_bu_av^b$.

Alternatively, you can think of $L^a_b$ as consuming a covector and outputting another covector $L^a_bu_a$. Or you can think of $L^a_b$ as consuming a vector and outputting another vector $L^a_bv^b$.

Similarly, you can think of a tensor with two upper indices such as $M^{bc}$ as a linear map from $T_p^* \times T_p^*$ to $\mathbb{R}$, or as a linear operator that:

(1) Consumes two covectors and outputs a scalar or

(2) Consumes one covector and outputs a vector