I've been studying representation theory of symmetric group on Tung's Group Theory in Physics. Theorem 13.14 discusses the the equivalence of irreducible representations of special linear group SL$(m)$ and their relation to those of general linear group GL$(m)$ via dual diagram. I understood that for general linear group, different Young Diagrams (on the space type $(i,j;k,l)$, namely, tensor spaces with undotted and dotted contravariant as well as covariant indices) corresponds to inequivalent irreducible representations (Theorem 13.12). Irreducible representation of SL$(m)$ can be obtained from those of GL$(m)$ by using a method involving invariant tensor $I$ defined and discussed in Lemma 2. In fact, two such invariant tensor exist, ${\delta_a}^b$ (Theorem 13.7) and $\varepsilon^{[a]}$ (Definition 13.10).
Since direct product of the above two types of tensor gives invariant tensor and therefore helps to construct the invariant tensor $I$ in Lemma 2. The procedure described in the textbook involves contract a set of $n$ anti-symmetric covariant indices, and the remaining $m-n$ contravariant indices will appear and be associated to the resulting new tensor. Though not stated explicitly in the textbook, it seems to me that this can be done by introducing direct product of $\delta$ invariant to those uncontracted indices so that the original tensor $T_{\tau(bc_1c_2\cdots)}^{\lambda(e)}$ can be transformed by the following invariant tensor $I$
$$\varepsilon^{ab}{\delta_{d_1}}^{c_1}{\delta_{d_2}}^{c_2}\cdots{\delta^{\lambda(e')}}_{\lambda(e)}= {(\varepsilon\delta\delta\cdots)^{a\lambda(e')}_{d_1d_2\cdots}}^{bc_1c_2\cdots}_{\lambda(e)} \to {(\varepsilon\delta\delta\cdots)^{\lambda'(ae')}_{\tau'(d_1d_2\cdots}}^{\tau(bc_1c_2\cdots)}_{\lambda(e)}$$
where $a$ and $b$ are subset of indices of the rank-$m$ anti-symmetric tensor, and the step involving "$\to$" is to properly symmetrize the indices so that they correspond to the correct Young Diagrams. The right hand side of the equation can be recognized as the invariant tensor introduced in Lemma 2:
$${I^{\lambda'(ae')}_{\tau'(d_1d_2\cdots)}}^{\tau(bc_1c_2\cdots)}_{\lambda(e)}$$
Such symmetric operation will not affect the invariant properly of the tensor. Now my questions are:
(1) How to show that $(\lambda,\tau)\to (\lambda',\tau')$ the rank of the tensor space does not change?
(2) Besides that $a$ and $b$ are anti-symmetric indices, why do they correspond to the same columns in the corresponding Young Diagrams?
Maybe some hits on the correct direction can be much appreciated.