Are we allowed to define a symmetric (1,1) tensor in the following way? I've recently stumbled accross the following task: "is it possible to define a symmetric and antisymmetric (1,1) tensor?". This is in context of relativity, so we have a metric $g$ at our disposal. Moreover, anything I say here should be applicable beyond the context of relativity, on any manifold equipped with a metric.
And we recall how (anti)symmetricity of a (0,2) tensor $M$ is defined. Two vectors $V$ and $W$ are considered and the following is demanded
$$
M (V, W) = \pm M (W, V)
$$
(plus for symmetric and minus for antisymmetric) or, in indices (by plugging in $\partial_\mu$ and $\partial_\nu$)
$$
M_{\mu\nu} = \pm M_{\nu\mu}
$$
Or similarly for tensor (2,0) we'd get $M^{\mu\nu} = \pm M^{\nu\mu}$ (in that case we are plugging in two one-forms)
Now I cannot do that with a tensor (1,1) because I cannot simply flip the arguments, but I can do the "next best thing" (hence, we start thinking about how we can define an analogous property for a mixed tensor). First, we take a covector $\tilde{A}$ and a vector $V$ and since we have a metric at our disposal, we can map the covector $\tilde{A}$ to a vector $A$ as $A = g^{-1} (\tilde{A}, \, \cdot \,)$ and map vector $V$ to covector as $\tilde{V} = g (V, \, \cdot \,)$, or in components
$$
\begin{aligned}
\tilde{A} &= A_\mu \mathrm{d} x^\mu \\
V &= V^\mu \partial_\mu
\end{aligned}
\;\,
\begin{aligned}
&\to \\
&\to
\end{aligned}
\quad
\begin{aligned}
A &= A^\mu \partial_\mu \\
\tilde{V} &= V_\mu \mathrm{d} x^\mu
\end{aligned}
$$
Now we can do the "next best thing" for a (1,1) mixed tensor and demand, that an "(anti)symmetric (1,1) tensor to satisfy"
$$
M (\tilde{A} ; V) = \pm M (\tilde{V} ; A)
$$
(again, plus for symmetric, minus for antisymmetric)
Which translates to
$$
M^\mu_{\;\; \nu} = \pm g^{\mu \rho} g_{\nu \sigma} M^\sigma_{\;\; \rho}
$$
and if we unwrap the $g$'s,
$$
M^\mu_{\;\; \nu} = \pm M_\nu^{\:\,\mu}
$$
which would be in a complete agreement with how a (2, 0) tensor $M^{\mu \nu}$ was symmetric, because if we take a tensor satisfying $M^{\mu \nu} = M^{\nu \mu}$ and simply drop an index by using the metric
$$
g_{\sigma \nu} M^{\mu \sigma} = g_{\sigma \nu} M^{\sigma \mu}
$$
we get the same answer
$$
M^\mu_{\;\; \nu} = M_\nu^{\:\,\mu}
$$
You can of course verify certain properties and little identities, that are analogous to (2,0) tensors.
For example, for a general (1,1) tensor, the (anti)symmetrization would correspond to
$$
(M^{(S)})^\mu_{\;\; \nu} = \frac{1}{2} \left( M^\mu_{\;\; \nu} + M_\nu^{\:\,\mu} \right) \quad \quad (M^{(A)})^\mu_{\;\; \nu} = \frac{1}{2} \left( M^\mu_{\;\; \nu} - M_\nu^{\:\,\mu} \right)
$$
Contracting a symmetric and antisymmetric tensor gives zero
$$
S^\mu_{\;\; \nu} A^\nu_{\;\,\mu} = 0
$$
Contracting a general and (anti)symmetric tensor only "reacts" with the (anti)symmetric part of the first tensor
$$
M^\mu_{\;\; \nu} S^\nu_{\;\; \mu} = (M^{(S)})^\mu_{\;\; \nu} S^\nu_{\;\; \mu} \quad \quad M^\mu_{\;\; \nu} A^\nu_{\;\; \mu} = (M^{(A)})^\mu_{\;\; \nu} A^\nu_{\;\; \mu}
$$
The "official" answer (book, professor, grader, ...) is that this is not possible and when I present this line of thinking I get a vague "you cannot do that" or answers that stray from my definitions (like "you cannot swap indices like that", which I would understand if I did $M^\mu_{\;\;\nu} + M^\nu_{\;\; \mu}$ however that's not what I'm doing here), or even answers that indicate that the person might not understand differential geometry, like "$M^\mu_{\;\; \nu}$ is identically equal to $M_\nu^{\;\mu}$, so what you call antisymmetric part, is always zero".
Which part of my reasoning is icky? The professor pointed out to me that in $M (\tilde{A} ; V) = M (\tilde{V} ; A)$ the objects fed to the tensor on the left-hand side are different from the objects fed to the tensor on the right-hand side, but that's part of my definition (*). Is there something logically wrong with this? I'd like someone versed in this topic to give me clear reasons why this is a no-no approach.
(*) And if this is the only objection, I find it rather weak. If this was the way people are thinking in mathematics, they would never define more general concepts stemming from simpler concepts, because they would never get past this kind of sentiment. A negative number? Have you ever seen minus two cows? A root of two? I've never had a root of two goats, what kind of nonsense is that? A factorial of a complex number? Come on, you cannot have $\pi + 2i$ numbers of people to divide $3-i$ marbles between, what will you come up with next? A fractional derivative? That doesn't make any sense! How do I write it as a d$y$/d$x$? I thought that in mathematics, we look for a useful way to generalize concepts that are already known to us, even if it means to go a bit beyond what seems "common sense" at the first sight.
Edit: I have two more ways to think about this now that I recall more details from the course of differential geometry.
First way, the metric provides a canonical isomorphism, so if we can define a concept of a symmetric (2,0) tensor, we can also define this concept on (1,1) tensors by mapping the corresponding (2,0) tensor to a (1,1) tensor by the musical isomorphism. Let $M$ be an (anti)symmetric tensor of rank (2,0), then (in indices), the corresponding (1,1) tensor is
$$
M^\mu_{\;\;\nu} = g_{\nu \sigma} M^{\mu \sigma}
$$
Second way, let $\left\langle \;, \; \right\rangle$ denote the inner product. Moreover, let's observe, that any tensor of the type (1,1) that has been fed a vector now provides a natural mapping from the cotangent space to $\mathbb{R}$,
$$
M : T^*M \otimes T M \to \mathbb{R} \quad \quad \implies M (\; ; V) : T^*M \to \mathbb{R}
$$
therefore, such object is a vector. We can then combine this with the inner product and define (anti)symmetric tensor of the rank (1,1) as follows
$$
\left\langle M ( \;\cdot\; ; V), W \right\rangle = \pm \left\langle V, M (\;\cdot\; ; W) \right\rangle
$$
This way, the objects $V$ and $W$ enter the equation in a very symmetric way, so the complaint from before does not hold.
This is, by the way, analogous to how we would conclude that a Laplacian is a "symmetric (1,1) tensor"/operator. First, a Laplacian acts on a function and spits out another function, so if we somehow understand functions to be "vectors" of a certain space, then Laplacian maps every vector to another vector, therefore, is a (1,1) tensor. The inner product can be defined as an integral, i.e.
$$
\left\langle \varphi, \psi \right\rangle = \int \mathrm{d}^3 x \, \varphi (x) \, \psi (x)
$$
and under certain conditions (we consider certain class of functions), the following is true for any two "vectors" $\varphi$, $\psi$ and the operator $\Delta$
$$
\left\langle \Delta \varphi, \psi \right\rangle = \left\langle \varphi, \Delta \psi \right\rangle
$$
 A: Yeah, it makes sense, but then $M$ is called "self adjoint" instead of "symmetric".
More precisely:
In finite dimension you have an isomorphism $\Phi: V^{**}\simeq V$ so, given the bilinear map $M:V^*\times V\to F$ you can get an endomorphism $\tilde M:V\to V$ given by $\tilde M(v) = \Phi(M({-},v))$ (and conversely from any endomorphism you can get a bilinear map, but I guess you know this).
Then the equation
$$\langle \tilde M(v),u\rangle = \langle v,\tilde M(u)\rangle$$
precisely says that $\tilde M$ is a self adjoint endomorpism, meaning that $\tilde M$ is self-adjoint with respect to the inner product $\langle-,-\rangle$.
It's a usual practice (as you yourself did) to conflate $\tilde M$ and $M$, and thus it is usual to say that $M$ itself is self adjoint.
In general, any linear endomorphism $A:V\to V$ in a finite-dimensional inner product space $V$ has a unique adjoint $B:V\to V$ such that
$$\langle A(v),u\rangle = \langle v,B(u)\rangle.$$
Also note that if you choose an orthonormal basis for $V$, then the matrices representing $A$ and $B$ with respect to that basis are tranposes of each other, but this need not be the case if the basis is not orthonormal.
