Order of arguments in a 1-1 tensor I was reading about staggered notation for tensor indices - and came across the fact that $T^b_{\ \ a}$ and $T^{\ \ b}_a$ may represent different tensors. $T^b_{\ \ a}=T(\varepsilon^b,e_a)$ and $T^{\ \ b}_a=T(e_a,\varepsilon^b)$. But I don't see why these two are different? A 1-1 tensor would just be a multilinear map on a vector and covector, and as far as I've read, no mention has been made about the order of the arguments.
Is it the case that $T^b_{\ \ a}$ is actually in $\mathcal{T}^1_1(V)$ and $T_a^{\ \ b}$ is in $\mathcal{T}^1_1(V^*)$ (and by convention we write the dual vector arguments first and the vector space arguments later) ? I'm not sure if this is the right reason and if not, I'd be grateful if the correct reason can be clarified.
 A: Let $V$ be an finite-dimensional vector space over $\Bbb{R}$. An $(r,s)$ tensor on $V$ is by definition an element of a vector space $W=W_1\otimes \cdots \otimes W_{r+s}$, such that $r$ of these factors are $V$ and the remaining $s$ factors are $V^*$. For the sake of concreteness, you can think $\underbrace{V\otimes\cdots \otimes V}_{\text{$r$ times}}\otimes \underbrace{V^*\otimes\cdots \otimes V^*}_{\text{$s$ times}}$. These can be identified with the space of $(r+s)$-multilinear maps $(V^*)^r\times (V)^s\to\Bbb{R}$. At this stage, the ordering is irrelevant (more on this below).
So, for example, a $(2,1)$ tensor on $V$ is by definition an element of $V\otimes V\otimes V^*$. But if you don’t like this ordering, you can also just all well call an element of $V^*\otimes V\otimes V$ a $(2,1)$ tensor.
A $(1,2)$ tensor is a completely different beast; it is an element of $V\otimes V^*\otimes V^*$ (someone else might come along and prefer a different way of ordering and use the space $V^*\otimes V^*\otimes V$… but that’s a minor difference). The ordering of the factors is irrelevant insofar as there is a canonical isomorphism which permutes the factors.

The above was just some general remarks about tensors. Now, to clarify your doubt, let us stick to $(1,1)$ tensors specifically, and use the multilinear maps language since that’s the interpretation you seem more comfortable with.
Let’s say I start with a multilinear map $F:V^*\times V\to\Bbb{R}$. I use $F$ because I don’t want to use $T$ a billion different times with different meanings (which is what everyone who knows what they’re talking about does).
Here’s a question: using the $F$, can you somehow construct a multilinear map $V\times V^*\to\Bbb{R}$? Yes, there is one very obvious candidate, namely $G:V\times V^*\to\Bbb{R}$, where I define
\begin{align}
G(v,\epsilon):=F(\epsilon,v).
\end{align}
Another way of writing what I’ve done is I used the “permuting isomorphism” $\pi: V\times V^*\to V^*\times V$, $\pi(v,\epsilon)=(\epsilon,v)$, and I defined
\begin{align}
G=F\circ \pi.
\end{align}
Very strictly speaking (and obviously), $F$ and $G$ are two very different maps (they have different domains), but this is just a simple reordering of the arguments (this is what I was referring to above when I said there is a canonical isomorphism which permutes the factors). So  in this sense, $F,G$ are different, but for silly/simple/technical reasons only.
If you write in components relative to a basis on $V$ and the dual basis on $V^*$, then we have $F^a_{\,\,b}=F(\epsilon^a,e_b)=G(e_b,\epsilon^a)=G_b^{\,\,a}$. Now, me personally, I don’t care too much about staggered vs non-staggered indices because I don’t work with indices too often anyway, and I know what I’m doing and I know what the abstract tensors are (which is why I gave different letters $F,G$).
But, the real reason why some people (eg physics) are so adamant about staggered vs non-staggered index placement is that in converting from the tensor $F:V^*\times V\to\Bbb{R}$ to a map $V\times V^*\to\Bbb{R}$, they don’t mean the obvious “permuted tensor” $G:=F\circ \pi$. They are talking about a completely separate beast! In the physics setting we often don’t just have an abstract vector space $V$ lying around, but we have a vector space $V$ together with a (pseudo)-inner product $g$ lying around (if $V=\Bbb{R}^3$, then $g$ is the usual inner/dot product, and in special relativity, $g$ is the Minkowski inner product, with your preferred signature). What this $g$ provides is an isomorphism $g^{\flat}:V\to V^*$, $g^{\flat}(v):=g(v,\cdot)$ (this requirement that this map be an isomorphism is the definition of $g$ being non-degenerate… note that positive-definiteness implies non-degeneracy). It is convention to call the inverse map $g^{\sharp}:V^*\to V$. See Wikipedia’s musical isomorphism. So, the availability of a $g$ allows you to convert vectors to covectors and covectors back to vectors (in general $V$ is isomorphic to $V^*$ because they have the same finite dimension, but there is no preferred/canonical isomorphism. If you have such a $g$, then there is a special isomorphism, which is the one I just defined).
Btw, if we’re talking about a single vector space $V$, then we might call this $g$ a “metric tensor on $V$”, whereas if you’re in the manifold setting you can do this at each tangent space so you have a metric tensor-field. Anyway, back to the main discussion. My setup now is I’m starting with a multilinear $F:V^*\times V\to\Bbb{R}$. I also have an isomorphism $g^{\flat}:V\to V^*$ with inverse $g^{\sharp}:V^{*}\to V$. Using this information, can I construct a bilinear map $V\times V^*\to\Bbb{R}$? One option was the $G=F\circ \pi$, but that we said was silly/trivial, so I want another one. Indeed, we can: define $\Phi:V\times V^*\to\Bbb{R}$ as
\begin{align}
\Phi(v,\epsilon):=F(g^{\flat}(v),g^{\sharp}(\epsilon)).
\end{align}
So, what I’m doing here is constructing an isomorphism $\mu:V\times V^*\to V^*\times V$ ($\mu$ for musical :) defined as $\mu(v,\epsilon)=(g^{\flat}(v),g^{\sharp}(\epsilon))$. Then, I’m defining $\Phi=F\circ \mu$.
Notice that this $\mu$ is not a simple permutation like the $\pi$ above, which is why $\Phi=F\circ \mu$ and $G=F\circ \pi$ are completely different guys even though they have the same domain and target space $V\times V^*\to\Bbb{R}$. An even more obvious way to see that $\Phi$ and $G$ are completely different beasts is that I constructed $G$ using only $F$ without any additional information, whereas to construct the $\Phi$, I needed $F$ and the additional data of a (pseudo)inner-product $g$.
Since $G,\Phi$ these are completely different multilinear maps (even though defined on the same domain), their components are obviously not necessarily equal: $\Phi(e_b,\epsilon^a)\neq G(e_b,\epsilon^a)$. Or, in staggered index notation (the staggering being there only to remind you which input comes from $V$ and which input comes from $V^*$),
\begin{align}
F^a_{\,\,b}=G_b^{\,\,a}\neq \Phi_{b}^{\,\,a}.
\end{align}
Now, the reason people make such a fuss about staggering indices and not mixing thing up is because

*

*They never ever mean the tensors obtained by a simple permutation of the factors. So, they’re talking about the pair $F,\Phi$, not the pair $F,G$. (Often in Physics people say the pair $F,\Phi$ are “physically related” by the metric duality… but this is just jargon you can ignore).

*They always use the same letter for the tensor, e.g $T$, and use the placement (up-down, and staggering) to keep track of which tensor specifically they’re talking about. So, rather than introducing two letters $F,\Phi$, they’ll simply say $T^a_{\,\,b}\neq T_b^{\,\,a}$.

Of course, point (2) has merits once you know what you’re talking about because then it becomes cumbersome to keep inventing new letters to talk about a different tensor when performing concrete computations. But if you’re just learning this stuff, then you should keep in mind that these are all different tensors (even though they are all $(1,1)$ tensors over $V$).
A: In first case you  have
$$T:V^*\times V\to\mathbb R$$
where you are evaluating $T(\varepsilon^i,e_j)$ to get the number $T^i{}_j$.
But the other case is
$$V\times V^*\to\mathbb R$$
where evaluation on $(e_i,\varepsilon^j)$ gives a number which not necessarily has to be $T^i{}_j$.
If you are employing the same letter to call this different maps then, maybe, this are causing the confusion.
